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

We begin with the given compound proposition

$$ (\sim p)\,\vee\,(p \wedge \sim q). $$

Our goal is to simplify it step by step, using the standard Boolean algebra (logical algebra) identities. The symbols have their usual meanings: $$\sim$$ denotes negation, $$\vee$$ denotes disjunction (OR), and $$\wedge$$ denotes conjunction (AND).

First, we recall the distributive law of logic, stated as

$$ A \;\vee\; (B \wedge C) \;\equiv\; (A \vee B)\; \wedge\; (A \vee C). $$

Here we can identify $$A = \sim p,$$ $$B = p,$$ and $$C = \sim q.$$ Substituting these into the distributive law, we obtain

$$ (\sim p) \;\vee\; (p \wedge \sim q) \;\equiv\; (\sim p \,\vee\, p)\; \wedge\; (\sim p \,\vee\, \sim q). $$

Next, notice that the expression $$\sim p \,\vee\, p$$ is a tautology because a statement OR its negation is always true. Symbolically,

$$ \sim p \,\vee\, p \;\equiv\; \text{True}. $$

When we have a conjunction in which one of the factors is the tautology, the whole conjunction reduces to the other factor. That is, $$\text{True} \wedge X \equiv X.$$ Therefore,

$$ (\sim p \,\vee\, p)\; \wedge\; (\sim p \,\vee\, \sim q) \;\equiv\; \text{True} \wedge (\sim p \,\vee\, \sim q) \;\equiv\; (\sim p \,\vee\, \sim q). $$

So we have shown

$$ (\sim p) \,\vee\, (p \wedge \sim q) \;\equiv\; (\sim p \,\vee\, \sim q). $$

Now we recognize another standard logical equivalence: the implication form. The implication

$$ p \to \sim q $$

is, by definition, equivalent to the disjunction

$$ \sim p \,\vee\, \sim q. $$

Since we have just derived the very same disjunction, it follows directly that

$$ (\sim p) \,\vee\, (p \wedge \sim q) \;\equiv\; p \to \sim q. $$

Thus, the original proposition is logically equivalent to the implication $$p \to \sim q,$$ which corresponds to Option A.

Hence, the correct answer is Option A.