The proposition $$\sim (p \vee \sim q) \vee \sim (p \vee q)$$ is logically equivalent to:

We start with the proposition: $$\sim (p \vee \sim q) \vee \sim (p \vee q)$$.

First, apply De Morgan's law to both parts. De Morgan's law states that $$\sim (A \vee B) = (\sim A) \wedge (\sim B)$$.

For the left part: $$\sim (p \vee \sim q)$$. Here, $$A = p$$ and $$B = \sim q$$, so:
$$\sim (p \vee \sim q) = (\sim p) \wedge \sim(\sim q)$$.
Since $$\sim(\sim q) = q$$, this simplifies to:
$$(\sim p) \wedge q$$.

For the right part: $$\sim (p \vee q)$$. Here, $$A = p$$ and $$B = q$$, so:
$$\sim (p \vee q) = (\sim p) \wedge (\sim q)$$.

Now the expression becomes:
$$(\sim p \wedge q) \vee (\sim p \wedge \sim q)$$.

Notice that $$\sim p$$ is common in both terms. Factor out $$\sim p$$:
$$\sim p \wedge (q \vee \sim q)$$.

We know that $$q \vee \sim q$$ is always true (tautology), because either $$q$$ is true or false. So:
$$q \vee \sim q = \text{True}$$.

Substitute this back:
$$\sim p \wedge \text{True}$$.

Any proposition AND True is the proposition itself. Therefore:
$$\sim p \wedge \text{True} = \sim p$$.

So, the original proposition simplifies to $$\sim p$$.

Comparing with the options:
A. p
B. q
C. $$\sim p$$
D. $$\sim q$$

Hence, the correct answer is Option C.