Let $$\triangle, \nabla \in \{\wedge, \vee\}$$ be such that $$(p \to q) \triangle (p \nabla q)$$ is a tautology. Then

We need to find $$\triangle, \nabla \in \{\wedge, \vee\}$$ such that $$(p \to q) \triangle (p \nabla q)$$ is a tautology.

Recall that $$p \to q \equiv \neg p \vee q$$.

Test all four combinations.

Case 1: $$\triangle = \vee, \nabla = \vee$$:

$$(\neg p \vee q) \vee (p \vee q) = \neg p \vee p \vee q = \text{T} \vee q = \text{T}$$

This is a tautology.

Case 2: $$\triangle = \vee, \nabla = \wedge$$:

$$(\neg p \vee q) \vee (p \wedge q)$$

When $$p = \text{T}, q = \text{F}$$: $$(\text{F} \vee \text{F}) \vee (\text{T} \wedge \text{F}) = \text{F} \vee \text{F} = \text{F}$$. Not a tautology.

Case 3: $$\triangle = \wedge, \nabla = \vee$$:

$$(\neg p \vee q) \wedge (p \vee q) = q$$

When $$q = \text{F}$$, the expression is $$\text{F}$$. Not a tautology.

Case 4: $$\triangle = \wedge, \nabla = \wedge$$:

$$(\neg p \vee q) \wedge (p \wedge q)$$

When $$p = \text{F}, q = \text{F}$$: $$\text{T} \wedge \text{F} = \text{F}$$. Not a tautology.

Conclusion.

Only $$\triangle = \vee, \nabla = \vee$$ gives a tautology.

The correct answer is Option C: $$\triangle = \vee, \nabla = \vee$$.