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

We have the compound proposition $$p \to (\sim p \vee \sim q).$$

First recall the logical rule for implication. For any two statements $$X$$ and $$Y$$, the implication $$X \to Y$$ is false only in the single case when the antecedent $$X$$ is true and the consequent $$Y$$ is false. Symbolically,

$$X \to Y = \text{False} \quad \Longleftrightarrow \quad X = \text{True} \ \text{and}\ Y = \text{False}.$$

In our problem, the antecedent is $$p$$ and the consequent is $$\sim p \vee \sim q.$$ The whole implication is declared to be false, so we must have

$$p = \text{True} \qquad\text{and}\qquad (\sim p \vee \sim q) = \text{False}.$$

Next consider the consequent, which is a disjunction (OR). For any two statements $$A$$ and $$B$$, the disjunction $$A \vee B$$ is false only when both $$A$$ and $$B$$ are false. Hence,

$$\sim p \vee \sim q = \text{False} \quad \Longleftrightarrow \quad \sim p = \text{False}\ \text{and}\ \sim q = \text{False}.$$

Now translate the falsity of the negations back to the original variables:

$$\sim p = \text{False} \quad \Longleftrightarrow \quad p = \text{True},$$

$$\sim q = \text{False} \quad \Longleftrightarrow \quad q = \text{True}.$$

We already had $$p = \text{True}$$ from the implication rule, and the analysis of the disjunction also forces $$p = \text{True}$$ and $$q = \text{True}.$$ Both requirements are consistent and give us the unique pair

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

Therefore, the truth values of $$p$$ and $$q$$ are respectively True and True, which corresponds to Option B.

Hence, the correct answer is Option B.