Above DFA can be redesigned as: [$S_0$ as $q00$, $S_1$ as $q10$ , $S_2$ as $q11$, $S_3$ as $q01$ ]
Where, each state as $q_{ab} [a = n_a \ mod \ 2 , b = n_b \ mod \ 2]$
$q00$ as $n_a \ mod \ 2 =0, n_b \ mod \ 2=0$ [no of $x$ is even no of $y$ is even ]
and $\delta(q00, x) \rightarrow q10 \ \ \ [(0+1) \ mod \ 2 =1$ as $x$ increase from $0$ to $1] \ \ \delta (q00, y) \rightarrow q01$
and $\delta (q10, x) \rightarrow q00 [(1+1) \ mod \ 2 =0]$ $\delta(q00, y) \rightarrow q01$ and soon
$q01$ is final state mean where no of $x$ is even and no of $y$ is odd
$q10$ is final state mean where no of $x$ is odd and no of $y$ is even.
So, D is correct answer.