Let $$r \in (P, q, \sim p, \sim q)$$ be such that the logical statement $$r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$$ is a tautology. Then $$r$$ is equal to

First, we need to find $$r \in \{p, q, \sim p, \sim q\}$$ such that $$r \vee (\sim p) \Rightarrow (p \wedge q) \vee r$$ is a tautology.

Next, recall that an implication $$P \Rightarrow Q$$ is a tautology when $$Q$$ is true whenever $$P$$ is true. Therefore, we need: whenever $$r \vee (\sim p)$$ is true, $$(p \wedge q) \vee r$$ must also be true.

Now, test $$r = \sim p$$. The statement becomes:

$$ (\sim p) \vee (\sim p) \Rightarrow (p \wedge q) \vee (\sim p) $$

Substituting, this simplifies to:

$$ \sim p \Rightarrow (p \wedge q) \vee (\sim p) $$

Since the implication must hold in all cases, consider the following:

Case 1: $$p$$ is True (so $$\sim p$$ is False). The implication $$F \Rightarrow \text{anything}$$ is True.

Case 2: $$p$$ is False (so $$\sim p$$ is True). The RHS becomes $$(F \wedge q) \vee T = T$$. So $$T \Rightarrow T$$ is True.

Since the implication is true in all cases, it is a tautology. ✓

Next, verify that the other options fail.

For $$r = p$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = q$$: Let $$p = F, q = F$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge F) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

For $$r = \sim q$$: Let $$p = T, q = T$$. Then LHS: $$F \vee F = F$$, so implication is True. Let $$p = F, q = T$$. Then LHS: $$F \vee T = T$$, RHS: $$(F \wedge T) \vee F = F$$. So $$T \Rightarrow F$$ is False. ✗

Therefore, the answer is Option C: $$\sim p$$.