For any two statement $$p$$ and $$q$$, the negative of the expression $$p \lor (\sim p \land q)$$ is:

We begin with the given propositional expression

$$E \;=\; p \,\lor\, (\sim p \,\land\, q).$$

Our task is to find the negation $$\sim E$$ and then match it with the choices. To do this comfortably, it is convenient first to simplify $$E$$ itself.

We recall the distributive law of propositional logic, stated as

$$A \,\lor\, (B \,\land\, C)\;=\; (A \,\lor\, B)\,\land\,(A \,\lor\, C).$$

Comparing, we see $$A = p,\; B = \sim p,\; C = q.$$ Substituting we obtain

$$p \,\lor\, (\sim p \,\land\, q) \;=\; (p \,\lor\, \sim p)\,\land\,(p \,\lor\, q).$$

Now, for any statement $$p$$, the expression $$p \,\lor\, \sim p$$ is always true; it is a tautology. We may therefore replace it by the propositional constant T (True):

$$ (p \,\lor\, \sim p)\,\land\,(p \,\lor\, q) \;=\; \text{T}\,\land\,(p \,\lor\, q).$$

The conjunction of a tautology with any statement leaves that statement unchanged, because T acts like the multiplicative identity in logic. Hence

$$E \;=\; p \,\lor\, q.$$

Having reduced the original expression to $$p \lor q,$$ we now negate it. For this we invoke De Morgan’s law, which states

$$\sim(A \,\lor\, B) \;=\; \sim A \,\land\, \sim B.$$

Applying the law directly with $$A = p$$ and $$B = q,$$ we get

$$\sim E \;=\; \sim(p \,\lor\, q) \;=\; \sim p \,\land\, \sim q.$$

This final form exactly matches Option C in the list.

Hence, the correct answer is Option C.