$L$ is Not regular and we can prove it using Pumping lemma for Regular languages.
Using Pumping lemma for Regular languages to Show that $L$ is Not Regular :
Pumping lemma states a Deep Property that All Regular languages share. By Showing that a language does not have the property stated by Pumping lemma, we are guaranteed that It is Not Regular.
Pumping Lemma(PL) is a necessary condition for Regular languages but not sufficient condition
i.e. All Regular languages must satisfy this condition and some Non-regular languages also satisfy this condition.
So, Regular Language → Satisfies Pumping Lemma
∴ Contrapositive statement is Doesn't Satisfy Pumping Lemma → Not Regular Language
Pumping lemma for Regular languages says
" If a language L is Regular,
then $\exists P \geq 1$, such that
$\forall \,\,strings\,\,w \in L$, If $|w| \geq P$ then
$\exists x,y,z$, such that $w = xyz$
and $|xy| \leq P$
and $y \neq \in$ i.e. $|y| \geq 1$
and $\forall q \geq 0$, $xy^qz \in L$
In Words, If $L$ is regular, then there’s some magic number $P$(Called Pumping length). And if I take any string $w$ that is at least as long as $P$, then I can break it up into three parts $x, y, $and $z.$ Now, the length of $xy$ is less than or equal to $P$, and also the length of $y$ is greater than or equal to $1$ and less than or equal to $P$.
In very simple words, If $L$ is a regular language then there is some positive number $P$ associated with $L$ such that for all strings $w $ of length greater than or equal to $P$, we can find some non-empty sub-string $y$ in $w$ within the first $P$ symbols of $w$ such that when we repeat $y$ Zero or more times then the produced strings also belong to $L.$
Now coming to the given language $L,$ Let's assume that it is Regular language. Hence, It will satisfy the Pumping lemma stated above.
So, There must be some positive number $P$ associated with $L$ such that Pumping lemma satisfies. So, Let's say that Magic number (Pumping length) is $P.$ So, Now, For all string, $w$ with length $\geq P$ , We must be able to find some Non-empty substring $y$ of length at most $P$ within the first $P$ symbols of $w$ such that we can repeat that $y$ Zero or more times, producing the strings belonging to $L.$
So, Let's take string $w = U^{P+1}D^{P},$ Clearly, $w \in L.$ For this string if You take any Non-empty substring $y$ within the first $P$ symbols (i.e. $y$ must be made of $U's$ only) and if you repeat that $y$ Zero times (repeating $y$ Zero times is same as Removing $y$ from the string $w$) then the resulted string will NOT belong to $L.$ So, Whatever Pumping length $P$ you take, at least One string $w = U^{P+1}D^{P} $ I have shown you which Won't get Pumped.
So, Pumping lemma is violated/Not Satisfied by $L.$ So, $L$ is Not Regular.
So, We have shown that $L$ is Not Regular.
Using Myhill-Nerode Equivalence Relation to Show that $L$ is Not Regular :
I will Not be stating the Whole Myhill-Neorde Theorem here But only the Part that is required to Show that $L_2'$ is Non-Regular.
Theorem : Let $L ⊆ Σ ^∗$ be any language. Suppose there is an infinite set $S$ of strings such that $\forall x, y ∈ S (x \neq y), x \not\equiv _L y.$ That is, suppose $(∀x \neq y ∈ S)(∃z ∈ Σ ^∗ ) (Exactly \,\,\,one\,\,\, of \,\,\,the \,\,\,strings\,\,\, xz\,\,\, or\,\,\, yz\,\,\, belongs\,\,\, to\,\,\, L).$ Then $L$ is not a regular language. Here such set $S$ is called Distinctive set for $L.$
http://www.cse.buffalo.edu/~regan/cse396/CSE396MNT.pdf
For given language $L, $ Consider the set $S = \{U^nD | n \geq 1\}.$ This set will work as a distinctive set for $L.$ i.e. If you take any Two different arbitrary string $x,y$ from $S$ then You will definitely find some $z \in \Sigma^*$ such that Exactly one of the Strings $xz$ or $yz$ will belong to $L.$ Let any pair $x, y ∈ S$ with $x \neq y $ be given. By definition of S, there are numbers $m, n ≥ 1$ with $m < n$ such that $x = U^mD$ and $y = U^nD$ . Take $z = D^{m-1}.$ Then $xz$ will Not belong to $L$ But $yz$ will belong to $L.$ Since $xz = U^mD^ m \notin L$ and $yz = U^nD^ m \in L(because $m <n$).$ Since $x, y ∈ S$ were arbitrary, $S$ is an “infinite distinctive set” for $L$, so $L$ is non-regular. (Here the boldface take goes with the $∃$ quantifer, while all the other boldface words go with the $∀$ quantifier.)
PDA for the Given Language $L$
CFG for modified $L$
I am going to give the CFG for slightly different language here.
Let $L'$ be the language of ALL Valid Strings i.e. I have removed the restriction of never returning to the base of the staircase after the final step. So, for this language $L'$ ,the CFG is as follows :
$S \rightarrow ASD \,\,|\,\,SS\,\,|\,\,A\,\,|\,\,\in$
$A \rightarrow UA \,\,|\,\,U$
Basic idea behind writing CFG for this language is that this language is somewhat similar to the language of Balanced parentheses.
The grammar for Balanced Parentheses language is this : $S \rightarrow (S) | SS|\in$
(Refer below for Balanced Parentheses Grammar and language :
https://cs.stackexchange.com/questions/14557/language-of-balanced-parentheses-biconditional-proof-about-parentheses )
So, The very basic idea here is that Every $D$ has some $U's(One \,\,or\,\, more)$ associated to it and those $U's$ appears before that $D$.
So, Just modify the Balanced Parentheses Grammar (Let me call it BP henceforth) to have any positive number of $U's$ associated with every $D's .$
So, The Simplest Grammar will come out to be :
$S \rightarrow ASD \,\,|\,\,SS\,\,|\,\,A\,\,|\,\,\in$
$A \rightarrow UA \,\,|\,\,U$
You can see that it is very similar to BP but it allows as to have more than one $U's $ associated with every $D.$
(I will leave it to the reader to convince himself/herself that this indeed is the Grammar associated with the modified language)
P.S. : I will update CFG for the given language in the Question as soon as I find it. If someone has any suggestions, Do let me know. I am kind of stuck because of the extra condition of "never returning to the base of the staircase after the final step. "
Maybe i am missing something trivial here or some small thing.