The Boolean Expression $$(p \wedge \sim q) \vee q \vee (\sim p \wedge q)$$ is equivalent to

We have to simplify the Boolean expression $$E=(p \wedge \sim q)\; \vee\; q\; \vee\; (\sim p \wedge q)$$ and compare the result with the four alternatives.

Because the $$\vee$$ (OR) and $$\wedge$$ (AND) operations are both commutative and associative, we may rearrange the terms without changing the value. Thus we write

$$E \;=\; q \;\vee\; (p \wedge \sim q) \;\vee\; (\sim p \wedge q).$$

Now notice the presence of the term $$q$$ by itself and the compound term $$(\sim p \wedge q).$$ The absorption law of Boolean algebra states that

$$x \;\vee\; (y \wedge x) \;=\; x.$$

Here, let us set $$x = q$$ and $$y = \sim p.$$ Applying the absorption law, we obtain

$$q \;\vee\; (\sim p \wedge q) = q.$$

Substituting this result back into the full expression gives

$$E \;=\; q \;\vee\; (p \wedge \sim q).$$

At this stage, only two terms remain. To proceed further we invoke the distributive law, which in Boolean form states

$$a \;\vee\; (b \wedge c) \;=\; (a \vee b) \;\wedge\; (a \vee c).$$

Assigning $$a = q,\; b = p,\; c = \sim q,$$ we have

$$q \;\vee\; (p \wedge \sim q)\;=\;(q \vee p)\;\wedge\;(q \vee \sim q).$$

The expression $$(q \vee \sim q)$$ is always true (a tautology), because either $$q$$ is true or $$\sim q$$ is true. In Boolean algebra, a tautology is represented by $$1$$. Therefore,

$$(q \vee p)\;\wedge\;1 \;=\; q \vee p.$$

Since the OR operation is commutative, $$q \vee p = p \vee q.$$ Hence the completely simplified form of the original expression is

$$E = p \vee q.$$

Looking at the options provided, this matches Option A.

Hence, the correct answer is Option A.