Negation of the Boolean expression $$p \leftrightarrow (q \rightarrow p)$$ is

We need to find the negation of the Boolean expression $$p \leftrightarrow (q \rightarrow p)$$.

$$q \rightarrow p \equiv \sim q \vee p$$

$$p \leftrightarrow (\sim q \vee p) \equiv [p \rightarrow (\sim q \vee p)] \wedge [(\sim q \vee p) \rightarrow p]$$

The first part: $$p \rightarrow (\sim q \vee p) \equiv \sim p \vee \sim q \vee p \equiv T$$ (tautology, since $$\sim p \vee p = T$$)

So: $$p \leftrightarrow (\sim q \vee p) \equiv T \wedge [(\sim q \vee p) \rightarrow p] \equiv (\sim q \vee p) \rightarrow p$$

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

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

So $$p \leftrightarrow (q \rightarrow p) \equiv p \vee q$$.

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

The negation is T only when both $$p$$ and $$q$$ are F, which is $$\sim p \wedge \sim q$$. $$\checkmark$$

Therefore, the correct answer is Option D: $$\sim p \wedge \sim q$$.