hello @srestha mam, Please check this link https://gateoverflow.in/93284/l-tm-tm-halts-on-every-input

In the given link, it is mentioned that :-

"$L = \left \{ \left \langle M \right \rangle | TM\ halts\ on\ every\ input\ \right \} \Rightarrow L = \left \{\left \langle M \right \rangle | L(M)\ is\ Recursive \ \right \}$"

Because a machine that halts for every input is called a Total Turing Machine and class of languages associated with these machines are called recursive languages.

Now in this question,

$1^{st}$ part of the problem says :- Turing machine $'M'$ halts on all inputs. So, language accepted by this Turing Machine '$M$' i.e. $L(M)$ Is recursive(decidable).

So, till now, we got a statement which says :- "$L(M)$ is recursive(decidable)."

Now, $2^{nd}$ part of the problem says :- "$L(M)=L'$ for some undecidable language $L'$"

According to this statement, $L(M)$ should be undecidable because it is equal to some undecidable language.

So, according to 2nd part :- "$L(M)$ is undecidable".

Now, there is an "**and**" between these 2 statements means both statement should be true simultaneously.

i.e. {$L(M)$ is recursive(decidable) and $L(M)$ is undecidable}

Since, it is not possible that L(M) is decidable and undecidable at the same time. So, no elements should occur in this set.

So, it should be empty set i.e. $\{\}$

I think instead of 'L' in LHS, they should have to give different letter because here if $L=\{\}=regular$ then its complement should also be regular which is contradicting by saying L' is undecidable. If we consider LHS as some set S then answer will be $S=\{\}$ which is regular set and also recursive(decidable) set.