Consider:
S1: $$p \Rightarrow q \lor p \land \sim q$$ is a tautology.
S2: $$\sim p \Rightarrow \sim q \land \sim p \lor q$$ is a contradiction.
Then

We need to check whether S1 is a tautology and S2 is a contradiction.

$$p \Rightarrow (q \lor p) \land (\sim q)$$. Actually, let us parse with standard precedence. $$\land$$ binds tighter than $$\lor$$, and $$\Rightarrow$$ has the lowest precedence. So the expression is:

$$p \Rightarrow (q \lor (p \land \sim q))$$

Let us evaluate using a truth table approach. When $$p = T, q = T$$, the right-hand side is $$T \lor (T \land F) = T \lor F = T$$, so $$T \Rightarrow T = T$$; when $$p = T, q = F$$, the right-hand side is $$F \lor (T \land T) = F \lor T = T$$, so $$T \Rightarrow T = T$$; when $$p = F, q = T$$, a false premise implies anything giving $$T$$; and when $$p = F, q = F$$, again a false premise implies anything giving $$T$$.

S1 is TRUE for all truth values, so S1 is a tautology. Statement S1 is correct.

$$\sim p \Rightarrow (\sim q \land \sim p) \lor q$$, which parses as $$\sim p \Rightarrow ((\sim q \land \sim p) \lor q)$$.

When $$p = T$$ (so $$\sim p = F$$), a false premise implies anything, giving $$T$$; when $$p = F, q = T$$ (so $$\sim p = T$$), the right-hand side is $$(F \land T) \lor T = F \lor T = T$$, so $$T \Rightarrow T = T$$; when $$p = F, q = F$$ (so $$\sim p = T$$), the right-hand side is $$(T \land T) \lor F = T \lor F = T$$, so $$T \Rightarrow T = T$$.

S2 is TRUE for all truth values, so S2 is also a tautology and not a contradiction, so the claim that S2 is a contradiction is wrong. Therefore, S1 is correct (it is a tautology), but S2 is incorrect (it is actually a tautology rather than a contradiction), and the correct answer is Option 4, for which only S1 is correct.