The logical statement $$[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r)$$ is equivalent to:

We have to simplify the statement $$[\sim(\sim p \vee q) \vee (p \wedge r)] \wedge (\sim q \wedge r).$$

First recall De Morgan’s law, which states $$\sim(A \vee B)\equiv (\sim A)\wedge(\sim B).$$ Applying this with $$A=\sim p$$ and $$B=q,$$ we obtain

$$\sim(\sim p \vee q)\equiv (\sim(\sim p))\wedge(\sim q)\equiv p \wedge \sim q.$$

Substituting this result back, the expression becomes

$$[(p \wedge \sim q) \vee (p \wedge r)] \wedge (\sim q \wedge r).$$

Now we focus on the disjunction inside the first bracket. Taking out the common factor $$p,$$ we use the distributive law $$X\wedge Y \,\vee\, X\wedge Z \equiv X\wedge(Y\vee Z).$$ Thus,

$$(p \wedge \sim q) \vee (p \wedge r)\equiv p \wedge (\sim q \vee r).$$

Hence the whole statement is now

$$\bigl[p \wedge (\sim q \vee r)\bigr] \wedge (\sim q \wedge r).$$

The connective $$\wedge$$ is associative and commutative, so we can rearrange terms without changing truth-value:

$$p \wedge (\sim q \vee r) \wedge \sim q \wedge r.$$

Observe that the factor $$\sim q \wedge r$$ already guarantees both $$\sim q$$ and $$r$$ are true, and whenever these two are true, the disjunction $$\sim q \vee r$$ is automatically true. Consequently the extra factor $$\sim q \vee r$$ is redundant inside the conjunction. Formally,

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

So the expression simplifies further to

$$p \wedge \sim q \wedge r.$$

Finally we may write this in the same order as option B:

$$(p \wedge r) \wedge \sim q.$$

This is exactly the statement given in option B. Hence, the correct answer is Option B.