$ \forall \left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}$ //Either P or Q should be true.

$ \forall x Q\left ( x \right )$ //Q is always true.

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$\therefore \exists x P\left ( x \right )$ // This means Q(x) is always True so P(x) can be true or false. So it is possible that $ \exists x P\left ( x \right )$ is True but we can't say surely about it.

so $A)\left \{ \forall \left \{ P\left ( x \right )\vee Q\left ( x \right ) \right \}, \forall x Q\left ( x \right ) \right \}\Rightarrow \exists x P\left ( x \right )$ is not valid but it is satisfiable.

$ \forall x P\left ( x \right ) $ //P is always true

$ \exists x \sim \left \{ P\left ( x \right ) \wedge Q\left ( x \right )\right \} $ \\ There exist a value of $x$ such that P and Q are not true at same time.

But we know P is always True. hence Q must be false for atleast one value of $x$.

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$\therefore \exists x \sim Q\left ( x \right )$ \\ There exist a value of $x$ for which Q is not true.

so $B) \left \{ \forall x P\left ( x \right ) , \exists x \sim \left \{ P\left ( x \right ) \wedge Q\left ( x \right )\right \} \right \} \Rightarrow \exists x \sim Q\left ( x \right )$ is valid