Let's analyze the given statement \( p \leftrightarrow (q \rightarrow p) \). This statement is a biconditional statement which is false only when one side is true and the other is false.
First, consider \( p \) as true and \( q \) as false. Then \( (q \rightarrow p) \) will be true, so \( p \leftrightarrow (q \rightarrow p) \) becomes true \( \leftrightarrow \) true, which is true. Thus, \( p \) true and \( q \) false doesn't make the given statement false.
Next, consider \( p \) as false and \( q \) as true. Then \( (q \rightarrow p) \) will be false, so \( p \leftrightarrow (q \rightarrow p) \) becomes false \( \leftrightarrow \) false, which is true. Therefore, \( p \) false and \( q \) true doesn't make the given statement false either.
Now, let's consider \( p \) as false and \( q \) as false. Then \( (q \rightarrow p) \) will be true. Therefore, \( p \leftrightarrow (q \rightarrow p) \) becomes false \( \leftrightarrow \) true, which is false. This satisfies the condition of the question.
Now, let's check the options for \( p \) as false and \( q \) as false:
A. \( p \) is false.
B. \( p \rightarrow (p \vee \sim q) \) becomes false \( \rightarrow \) (false \( \vee \) true), which is false \( \rightarrow \) true = true.
C. \( p \wedge (\sim p q) \) becomes false \( \wedge \) (true \( \wedge \) false) = false.
D. \( (p \vee \sim q) \rightarrow p \) becomes (false \( \vee \) true) \( \rightarrow \) false = true \( \rightarrow \) false = false.
Option B is the only statement which is true when \( p \) is false and \( q \) is false.