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

We begin with the expression $$\sim(\sim p \to q)$$ and shall simplify it carefully, showing every algebraic detail.

First, we recall the implication law:

$$x \to y \;\equiv\; \sim x \;\vee\; y.$$

Here the part playing the role of $$x$$ is $$\sim p$$ and the part playing the role of $$y$$ is $$q$$. Substituting these symbols into the implication law we obtain

$$\bigl(\,\sim p \to q\,\bigr) \;\equiv\; \sim(\,\sim p\,)\;\vee\;q.$$

Now, inside the negation we have a double negation $$\sim(\,\sim p\,)$$, and the rule of double negation states that $$\sim(\,\sim p\,)\equiv p$$. Hence we get

$$\bigl(\,\sim p \to q\,\bigr) \;\equiv\; p \;\vee\; q.$$

Returning to the original expression, we must negate this result. So we write

$$\sim(\,\sim p \to q\,) \;\equiv\; \sim\bigl(p \;\vee\; q\bigr).$$

Next, we invoke De Morgan’s law, which tells us that the negation of a disjunction is the conjunction of the negations:

$$\sim\bigl(p \;\vee\; q\bigr) \;\equiv\; \bigl(\sim p\bigr) \;\wedge\; \bigl(\sim q\bigr).$$

Thus, after carrying out all transformations, the original expression simplifies completely to

$$\sim p \;\wedge\; \sim q.$$

Looking at the options, this matches Option B.

Hence, the correct answer is Option B.