The statement $$(\sim(p \Leftrightarrow \sim q)) \wedge q$$ is:

We need to determine what the statement $$(\sim(p \Leftrightarrow \sim q)) \wedge q$$ is equivalent to.

$$p \Leftrightarrow \sim q$$ is true when p and $$\sim q$$ have the same truth value, i.e., when p and q have opposite truth values.

Therefore, $$\sim(p \Leftrightarrow \sim q)$$ is true when p and q have the same truth value, which means $$\sim(p \Leftrightarrow \sim q) \equiv (p \Leftrightarrow q)$$.

The statement becomes $$(p \Leftrightarrow q) \wedge q$$.

$$p = T, q = T$$: $$(T \Leftrightarrow T) \wedge T = T \wedge T = T$$

$$p = T, q = F$$: $$(T \Leftrightarrow F) \wedge F = F \wedge F = F$$

$$p = F, q = T$$: $$(F \Leftrightarrow T) \wedge T = F \wedge T = F$$

$$p = F, q = F$$: $$(F \Leftrightarrow F) \wedge F = T \wedge F = F$$

The statement is true only when $$p = T$$ and $$q = T$$, which is $$p \wedge q$$.

Option D: $$(p \Rightarrow q) \wedge p$$

$$p = T, q = T$$: $$T \wedge T = T$$

$$p = T, q = F$$: $$F \wedge T = F$$

$$p = F, q = T$$: $$T \wedge F = F$$

$$p = F, q = F$$: $$T \wedge F = F$$

This is also equivalent to $$p \wedge q$$. ✓

Therefore, the correct answer is Option D: equivalent to $$(p \Rightarrow q) \wedge p$$.