Consider :
Statement - I : $$(p \wedge \sim q) \wedge (\sim p \wedge q)$$ is a fallacy.
Statement - II : $$(p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p)$$ is a tautology.

First we recall some basic logical equivalences that we shall use again and again.

Implication : $$\, (A \rightarrow B)\; \equiv\; (\sim A)\,\vee\,B \,.$$ Biconditional : $$\, (X \leftrightarrow Y)\; \equiv\; (X \wedge Y)\,\vee\,(\sim X \wedge \sim Y)\;.$$ Associative and commutative laws allow us to rearrange and regroup the symbols $$\wedge$$ and $$\vee$$ freely.

Now we examine Statement I :

We have $$ (p \wedge \sim q) \wedge (\sim p \wedge q). $$ Because the conjunction $$\wedge$$ is associative, the brackets may be dropped one at a time:

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

Next, with the commutative law we collect like symbols together:

$$ p \wedge \sim p \wedge \sim q \wedge q. $$

Inside the expression we clearly see $$p \wedge \sim p$$ and also $$q \wedge \sim q$$. Each of these pairs is always false because a statement and its negation can never be true simultaneously. Hence

$$ p \wedge \sim p \;=\; \text{False}, \qquad q \wedge \sim q \;=\; \text{False}. $$

So the whole conjunction becomes

$$ \text{False} \wedge \text{False} \;=\; \text{False}. $$

The result does not depend on the particular truth-values of $$p$$ or $$q$$; it is always false. A statement that is always false is called a fallacy. Therefore Statement I is true.

Next we analyse Statement II :

We start with the given biconditional $$ (p \rightarrow q) \leftrightarrow (\sim q \rightarrow \sim p). $$

Using the implication formula quoted at the beginning, we transform each part separately.

First, $$ p \rightarrow q \;\equiv\; \sim p \,\vee\, q. $$

Second, we handle the contrapositive implication carefully. Applying $$(A \rightarrow B) \equiv (\sim A) \vee B$$ with $$A = \sim q$$ and $$B = \sim p$$ we obtain

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

Notice that the disjunction $$q \vee \sim p$$ is identical to $$\sim p \vee q$$ because $$\vee$$ is commutative. Hence we can write

$$ q \vee \sim p \;=\; \sim p \vee q. $$

So both sides of the biconditional are in fact the same formula:

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

Therefore the whole statement becomes

$$ (\sim p \vee q) \leftrightarrow (\sim p \vee q). $$

A biconditional of any statement with itself, i.e. $$X \leftrightarrow X,$$ is always true, because both possible parts in the definition $$(X \wedge X) \vee (\sim X \wedge \sim X)$$ are automatically satisfied. Hence the biconditional above is always true; that is, it is a tautology. So Statement II is also true.

We must still decide whether Statement II provides a correct explanation for Statement I. Statement II merely tells us that an implication is equivalent to its contrapositive; it does not explain why the particular conjunction in Statement I is always false. Thus, while both statements are true, Statement II is not the reason for Statement I.

The situation matches Option D.

Hence, the correct answer is Option D.