The Boolean expression $$\sim(p \Rightarrow (\sim q))$$ is equivalent to

We have to simplify the Boolean expression $$\sim\!\bigl(p \Rightarrow (\sim q)\bigr)$$ and write the result in one of the listed equivalent forms.

First, we recall the logical equivalence that defines an implication. The implication $$a \Rightarrow b$$ is always equivalent to the disjunction $$(\sim a)\,\vee\, b.$$ We will apply this fact to the inner implication $$p \Rightarrow (\sim q).$$

So we write

$$p \Rightarrow (\sim q) \;=\; (\sim p)\,\vee\,(\sim q).$$

Substituting this back into the original expression gives

$$\sim\!\bigl(p \Rightarrow (\sim q)\bigr) \;=\; \sim\!\bigl((\sim p)\,\vee\,(\sim q)\bigr).$$

Now we must remove the outer negation. For this we make use of De Morgan’s Law, which states:

$$\sim(A \vee B) = (\sim A) \wedge (\sim B), \quad \sim(A \wedge B) = (\sim A) \vee (\sim B).$$

In our case $$A = (\sim p)\quad\text{and}\quad B = (\sim q).$$ Hence

$$\sim\!\bigl((\sim p)\,\vee\,(\sim q)\bigr) = (\sim(\sim p)) \,\wedge\, (\sim(\sim q)).$$

Double negations cancel out, because $$\sim(\sim r)=r$$ for any statement $$r$$. Therefore

$$ (\sim(\sim p)) \,\wedge\, (\sim(\sim q)) = p \,\wedge\, q.$$

Thus we have shown step by step that

$$\sim\!\bigl(p \Rightarrow (\sim q)\bigr) = p \wedge q.$$

The equivalent option in the list is Option D.

Hence, the correct answer is Option D.