If $$p \to (p \wedge \sim q)$$ is false, then the truth values of $$p$$ and $$q$$ are respectively:

We are told that the proposition $$p \to (p \wedge \sim q)$$ is false.

First, we recall the fundamental rule for a material implication. The compound statement $$A \to B$$ is false only in the single case when the antecedent $$A$$ is true and the consequent $$B$$ is false. Symbolically,

$$A \to B = F \;\; \text{iff} \;\; A = T \text{ and } B = F.$$

In our problem, the antecedent is $$p$$ and the consequent is $$p \wedge \sim q.$$ So, for the whole implication to be false, we must have

$$p = T \quad\text{and}\quad (p \wedge \sim q) = F.$$

We already know from the first part that

$$p = T.$$

Now we must analyze the conjunctive statement $$p \wedge \sim q.$$ A conjunction $$X \wedge Y$$ is true only when both $$X$$ and $$Y$$ are true; otherwise it is false. Here,

$$X = p, \qquad Y = \sim q.$$

We have just found $$p = T.$$ Let us substitute this known value:

$$p \wedge \sim q = T \wedge \sim q.$$

We are told that this entire conjunction is actually false:

$$T \wedge \sim q = F.$$

The only way an expression of the form $$T \wedge Y$$ can be false is when the second factor $$Y$$ is false. Hence,

$$\sim q = F.$$

A negation $$\sim q$$ is false precisely when the original statement $$q$$ is true. Therefore,

$$q = T.$$

Collecting our results, we have obtained

$$p = T, \qquad q = T.$$

This matches Option C, which lists the truth values as $$T, T.$$

Hence, the correct answer is Option 3.