Which of the following statement is a tautology?

We need to identify which of the given statements is a tautology (always true regardless of truth values of $$p$$ and $$q$$).

Analyze Option A: $$((\sim q) \wedge p) \wedge q$$

This simplifies to $$p \wedge q \wedge (\sim q)$$, which contains $$q \wedge (\sim q) = F$$.

So this is always False (contradiction). Not a tautology.

Analyze Option B: $$((\sim q) \wedge p) \wedge (p \wedge (\sim p))$$

The term $$p \wedge (\sim p) = F$$.

So the entire expression is $$(\text{anything}) \wedge F = F$$.

This is always False (contradiction). Not a tautology.

Analyze Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$

The term $$p \vee (\sim p) = T$$ (this is a tautology by itself).

So the entire expression is $$(\text{anything}) \vee T = T$$.

This is always True. This is a tautology!

Analyze Option D: $$(p \wedge q) \wedge (\sim(p \wedge q))$$

Let $$r = p \wedge q$$. Then this becomes $$r \wedge (\sim r) = F$$.

This is always False (contradiction). Not a tautology.

Therefore, the tautology is Option C: $$((\sim q) \wedge p) \vee (p \vee (\sim p))$$.

The correct answer is Option C.