Eliminate each option by throwing a counter example
L – Context free language, M – Regular language
A) always regular
$L$ - {$a^{n}b^{n}$}
$R$ - {$a^{*}b^{*}$}
$L \cap R$ = {$a^{n}b^{n}$}, which is not regular
Therefore, A) is False
B) Never regular
$L$ = { }
$R$ = { }
$L \cap R$ = { }, which is regular
Therefore B) is False
C) always a deterministic context-free language
$L$ - $\left \{ ww^{r}, w\in \left \{ a,b \right \}^{*} \right \}$
$R$ - $\left \{ (a+b)^{*} \right \}$
$L \cap R$ =$\left \{ ww^{r}, w\in \left \{ a,b \right \}^{*} \right \}$, which is non-deterministic CFL
Therefore, C) is False
So, Option D is True
Actually, to come to a conclusion of “always correct”, you need to come up with the proof. You can’t just show one example as “correct” and come to a conclusion of “always correct”.
But you can come to a conclusion of “not always correct” by just throwing a counter example which tells as “not correct”.