Answer : B
$A = \left \{ xy | x,y \in \left \{ a,b \right \}^*, n_a(x) = n_b(y) \right \}$
It is Regular. Moreover, $A = \Sigma^*$
Let's Prove it Formally.
Proof :
We can see easily(check manually) that $\in, a,b,aa,ab,ba,bb$ etc are in the language $A$ (i.e. We can split all these Strings in Two parts such that in the left part there are as many $a's$ as there are $b's$ in right part. Just check manually)
Two length strings can be partitioned as follows : $\left \{ |aa, a|b,b|a,bb| \right \}$
Now, All the Three length strings can be formed by appending $a$ or $b$ to the Two length strings and We already have some partition of Two length strings. So, Here goes the Idea :
"If we append $a$ at the end of any Two length string then the Previous Partition of the Two length string will remain as it is"
i.e. $\left \{ |aaa, a|ba,b|aa,bb|a \right \}$
"If we append $b$ at the end of any Two length string then the Previous Partition of the Two length string will shift to right by One Position"
i.e. $\left \{ a|ab, ab|b,ba|b,bbb| \right \}$
So, We can inductively say that ""If we append $a$ at the end of any $n-1$ length string then the Previous Partition of the $n-1$ length string will remain as it is" and "If we append $b$ at the end of any $n-1$ length string then the Previous Partition of the $n-1$ length string will shift to right by One position"
Hence, We can say that Every string can be Partitioned in Two parts such that in the left part there are as many $a's$ as there are $b's$ in right part. So, $A = \Sigma^*$
$B$ is Simple and Standard Non-regular CFL.