If the Boolean expression $$(p \wedge q) \circledast (p \otimes q)$$ is a tautology, then $$\circledast$$ and $$\otimes$$ are respectively given by:

We need to find the connectives $$\circledast$$ and $$\otimes$$ such that $$(p \wedge q) \circledast (p \otimes q)$$ is a tautology.

Let us check Option A: $$\circledast$$ is $$\to$$ and $$\otimes$$ is $$\to$$. The expression becomes $$(p \wedge q) \to (p \to q)$$.

Recall that an implication $$A \to B$$ is false only when $$A$$ is true and $$B$$ is false. So $$(p \wedge q) \to (p \to q)$$ is false only when $$(p \wedge q)$$ is true and $$(p \to q)$$ is false.

If $$(p \wedge q)$$ is true, then both $$p$$ and $$q$$ are true. But when $$p$$ is true and $$q$$ is true, $$(p \to q)$$ is also true. So the antecedent being true forces the consequent to be true as well.

Therefore $$(p \wedge q) \to (p \to q)$$ can never be false, making it a tautology.

Let us verify the other options fail. For Option B ($$\circledast = \wedge$$, $$\otimes = \vee$$): the expression is $$(p \wedge q) \wedge (p \vee q)$$. When $$p = T, q = F$$: $$(T \wedge F) \wedge (T \vee F) = F \wedge T = F$$. Not a tautology.

For Option C ($$\circledast = \vee$$, $$\otimes = \to$$): the expression is $$(p \wedge q) \vee (p \to q)$$. When $$p = T, q = F$$: $$(T \wedge F) \vee (T \to F) = F \vee F = F$$. Not a tautology.

For Option D ($$\circledast = \wedge$$, $$\otimes = \to$$): the expression is $$(p \wedge q) \wedge (p \to q)$$. When $$p = F, q = F$$: $$(F \wedge F) \wedge (F \to F) = F \wedge T = F$$. Not a tautology.

The answer is $$\to, \to$$, which is Option A.