It's simply based on Demorgan's law
$\sim \exists x(\forall y(\alpha )\wedge \forall z(\beta ))$
// Simply Replace $\forall$ with $\exists$ and vice versa , $\vee$ with $\wedge$ and vice versa , $P(t)$ with $\sim$$P(t)$ and vice versa.
$\equiv$ $\forall x(\exists y(\sim\alpha )\vee\exists z(\sim(\beta )))$
// like $\overline{a+b}\equiv \overline{a} .\overline{b}$ this $\overline{\forall xP(t)} \equiv \exists x\sim P(t)$
$\equiv \forall x(\forall z(\beta )\rightarrow \exists y(\sim(\alpha )))$ -------option B matched
Let suppose $\forall z(\beta )$ is 'a' and $\exists y(\sim\alpha )$ is 'b'
So option B is like $a->b$ and option C is $\sim b -> \sim a$ , both are equivalent so hence C is also equivalent.
Option A is $\sim a -> \sim b$ which is not equivalent to $a->b$ so out
Option D is $b->\sim a$ which is also not equivalent so out.
