$\sim ∀ x$ $[P(x) \rightarrow (Q(x) \vee P(x))]$
this can also be written as
$\sim$ ∀x$[(P(x) \rightarrow (Q(x) )\vee (P(x)\rightarrow P(x) )]$
$P(x)\rightarrow P(x)$ will alwaysb be true,
$\sim$ ∀x$[(P(x) \rightarrow (Q(x) )\vee True]$
$\sim$ ∀x[True]
"This is not the case that for all "x" the expression is True"
We know that, $\sim$∀$x P(x)$ $\equiv ∃x \sim P(x)$
$∃x \sim[True]$
$∃x [False]$
Value of predicate which is false does not depends on what is the value of x. Hence the expression is a Contradiction.