For brevity, I will write $P(x)$ as $P_x$
Let the domain of $x = \{1, 2\}$
Option A:
$(\forall x P(x) \lor \forall x Q(x)) \rightarrow \forall x (P(x) \lor Q(x))$
$=(P_1P_2 + Q_1Q_2) \rightarrow (P_1 + Q_1)(P_2 + Q_2)$
$=(P_1P_2 + Q_1Q_2) \rightarrow (P_1P_2 + Q_1Q_2 + P_1Q_2 + P_2Q_1)$
$=(P_1P_2 + Q_1Q_2)’ + (P_1P_2 + Q_1Q_2) + (P_1Q_2 + P_2Q_1)$
$=true$
$\therefore$ valid
Option B:
$(\exists x P(x) \land \exists x Q(x)) \rightarrow \exists x (P(x) \land Q(x))$
$=\{(P_1 + P_2)(Q_1 + Q_2)\} \rightarrow (P_1Q_1 + P_2Q_2)$
$=\{(P_1 + P_2)(Q_1 + Q_2)\}’ + (P_1Q_1 + P_2Q_2)$
$=P_1’P_2’ + Q_1’Q_2’ + P_1Q_1 + P_2Q_2$
$\therefore$ NOT valid
Option C:
$\forall x (P(x) \lor Q(x)) \rightarrow (\forall x P(x) \lor \forall x Q(x))$
$= \{(P_1 + Q_1)(P_2 + Q_2)\} \rightarrow (P_1P_2 + Q_1Q_2)$
$= \{(P_1 + Q_1)(P_2 + Q_2)\}’ + (P_1P_2 + Q_1Q_2)$
$= P_1’Q_1’ + P_2’Q_2’ + P_1P_2 + Q_1Q_2$
$\therefore$ NOT valid
Option D:
$\exists x (P(x) \lor Q(x)) \rightarrow \neg (\forall x P(x) \lor \exists x Q(x))$
$=\exists x (P(x) \lor Q(x)) \rightarrow (\exists x \neg P(x) \land \forall x \neg Q(x))$
$= \{(P_1 + Q_1)(P_2 + Q_2)\} \rightarrow (P_1’ + P_2’)(Q_1’Q_2’)$
$= P_1’Q_1’ + P_2’Q_2’ + (P_1’ + P_2’)(Q_1’Q_2’)$
$=P_1’Q_1’ + P_2’Q_2’$
$\therefore$ NOT valid