Let's analyze the statement \( p \leftrightarrow (q \rightarrow p) \). This is a biconditional statement, which is false only when its components have differing truth values.Case 1: \( p \) is true, \( q \) is false. \( (q \rightarrow p) \) is true. Thus, \( p \leftrightarrow (q \rightarrow p) \) becomes true \( \leftrightarrow \) true, which is true. This case does not falsify the statement.Case 2: \( p \) is false, \( q \) is true. \( (q \rightarrow p) \) is false. Thus, \( p \leftrightarrow (q \rightarrow p) \) becomes false \( \leftrightarrow \) false, which is true. This case also does not falsify the statement.Case 3: \( p \) is false, \( q \) is false. \( (q \rightarrow p) \) is true. Thus, \( p \leftrightarrow (q \rightarrow p) \) becomes false \( \leftrightarrow \) true, which is false. This case satisfies the condition of falsifying the statement.Now, we evaluate the given options for \( p \) false and \( q \) false:A. \( p \) is false.B. \( p \rightarrow (p \vee \sim q) \) becomes false \( \rightarrow \) (false \( \vee \) true) which evaluates to false \( \rightarrow \) true, resulting in true.C. \( p \wedge (\sim p q) \) becomes false \( \wedge \) (true \( \wedge \) false) which evaluates to false \( \wedge \) false, resulting in false.D. \( (p \vee \sim q) \rightarrow p \) becomes (false \( \vee \) true) \( \rightarrow \) false which evaluates to true \( \rightarrow \) false, resulting in false.Option B is the only statement that is true when \( p \) is false and \( q \) is false.