The boolean expression $$(\sim(p \wedge q)) \vee q$$ is equivalent to

We need to find the expression equivalent to $$(\sim(p \wedge q)) \vee q$$.

Using De Morgan's law: $$\sim(p \wedge q) = \sim p \vee \sim q$$. Substituting gives $$(\sim(p \wedge q)) \vee q = (\sim p \vee \sim q) \vee q = \sim p \vee (\sim q \vee q) = \sim p \vee T = T$$ (since $$\sim q \vee q$$ is always true), so the original expression is a tautology.

Next, we examine each option:

Option A: $$q \to (p \wedge q) = \sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q) = (\sim q \vee p) \wedge T = \sim q \vee p$$. This is NOT always true (fails when $$p = F, q = T$$), so it is not a tautology.

Option B: $$p \to q = \sim p \vee q$$. This is NOT always true (fails when $$p = T, q = F$$), so it is not a tautology.

Option C: $$p \to (p \to q) = p \to (\sim p \vee q) = \sim p \vee (\sim p \vee q) = \sim p \vee q$$. This is the same as Option B and thus not a tautology.

Option D: $$p \to (p \vee q) = \sim p \vee (p \vee q) = (\sim p \vee p) \vee q = T \vee q = T$$. This is always true — a tautology.

Since the given expression simplifies to a tautology, and Option D is also a tautology, they are logically equivalent.

The correct answer is Option D.