The negation of the Boolean expression $$p \vee (\sim p \wedge q)$$ is equivalent to:

We are asked to find the negation of the Boolean expression $$p \vee (\sim p \wedge q)$$ and then match the simplified result with one of the given options.

First we write the required negation explicitly:

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

We recall De Morgan’s first law, which states that for any two propositions $$A$$ and $$B$$, the negation of their disjunction is

$$\neg(A \vee B)=\neg A \wedge \neg B.$$

In our expression, we can treat $$A$$ as $$p$$ and $$B$$ as $$(\sim p \wedge q)$$. Applying the law we get

$$\neg\bigl[p \vee (\sim p \wedge q)\bigr] \;=\; \bigl(\neg p\bigr) \;\wedge\; \neg\bigl(\,\sim p \wedge q\,\bigr).$$

Now we turn to the second negation inside. Again, by De Morgan’s second law, which says

$$\neg(X \wedge Y)=\neg X \vee \neg Y,$$

we let $$X=\sim p$$ and $$Y=q$$. Thus

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

The double negation rule $$\neg(\sim p)=p$$ simplifies the first term, so we have

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

Substituting this back into the earlier expression yields

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

We now use the distributive law of conjunction over disjunction, namely

$$X \wedge (Y \vee Z) = (X \wedge Y) \vee (X \wedge Z).$$

Here $$X=\neg p,$$ $$Y=p,$$ and $$Z=\neg q.$$ Distributing gives

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

Observing that $$\neg p \wedge p$$ is always false (a contradiction), we replace it by the constant false value, traditionally denoted by $$0$$. Since $$0 \vee$$ (anything) is just that “anything,” the expression simplifies to

$$\neg p \wedge \neg q.$$

Thus the negation of $$p \vee (\sim p \wedge q)$$ is exactly $$\sim p \wedge \sim q$$, which corresponds to Option B.

Hence, the correct answer is Option B.