If q is false and $$p \wedge q \leftrightarrow r$$ is true, then which one of the following statements is a tautology?

We are told that the statement $$q$$ is false and at the same time the biconditional $$(p \wedge q) \leftrightarrow r$$ is true. Let us first interpret this information.

By definition, the conjunction $$p \wedge q$$ is true only when both $$p$$ and $$q$$ are true. Because we already know that $$q$$ is false, the value of the entire conjunction is immediately fixed:

$$p \wedge q = \text{False}.$$

Next, we look at the biconditional. Remember the rule for a biconditional:

For any two statements $$A$$ and $$B$$, the statement $$A \leftrightarrow B$$ is true precisely when either
(i) both $$A$$ and $$B$$ are true, or
(ii) both $$A$$ and $$B$$ are false.

In our problem $$A = (p \wedge q)$$ and $$B = r$$. We have discovered that $$A$$ is false. For the biconditional $$A \leftrightarrow B$$ to be true, case (ii) must hold; therefore $$B$$ must also be false. Concretely,

$$r = \text{False}.$$

At this point we have

$$q = \text{False}, \quad r = \text{False}, \quad p$$ is still free (it can be True or False).

We now inspect each option to see which statement is necessarily true—i.e. a tautology—under these fixed truth values.

Option A: $$(p \vee r) \to (p \wedge r).$$
Because $$r = \text{False},$$ we simplify step by step:

$$p \vee r = p \vee \text{False} = p,$$
$$p \wedge r = p \wedge \text{False} = \text{False}.$$

Thus Option A becomes $$p \to \text{False}.$$ The truth rule for implication says $$A \to B$$ is false exactly when $$A$$ is true and $$B$$ is false. Here, whenever $$p$$ is true the implication is false. Hence Option A is not always true.

Option B: $$(p \wedge r) \to (p \vee r).$$
Again using $$r = \text{False}$$ we get

$$p \wedge r = p \wedge \text{False} = \text{False},$$
$$p \vee r = p \vee \text{False} = p.$$

The whole statement reduces to $$\text{False} \to p.$$ An implication whose antecedent is false is always true, irrespective of the consequent. Therefore Option B is always true for both possibilities of $$p$$ and is thus a tautology.

Option C: $$p \wedge r = p \wedge \text{False} = \text{False}.$$ This is always false, so it certainly is not a tautology.

Option D: $$p \vee r = p \vee \text{False} = p.$$ The truth of this statement depends on whether $$p$$ itself is true or false, so it is not guaranteed to be true in all cases.

Only Option B satisfies the requirement of being always true under the given conditions.

Hence, the correct answer is Option B.