$X \equiv (P \vee Q) → R$
$\quad \equiv \neg(P \vee Q) \vee R$
$\quad \equiv (\neg P \wedge \neg Q) \vee R$
$\quad \equiv (\neg P \vee R) \wedge (\neg Q \vee R)$
$\quad \equiv (P → R) \wedge (Q → R)$
So, $X → Y$ is true as $(A \wedge B) → (A \vee B)$ is always TRUE but reverse implication is not always true.
Hence, B.