No valid argument can prove this. Suppose $M, N, S$ and $W$ are true, and $A$ and $B$ are false. Then the three premisses are all true, but the conclusion is false.
\begin{align} \lnot p \vee q \rightarrow r \tag{1}\\
s \vee \lnot q \tag{2} \\
\lnot t \tag{3}\\
p \rightarrow t \tag{4}\\
\lnot p \wedge r \rightarrow \lnot s \tag{5}\end{align}
As you note in your post, from the fourth premise we have $p\to t$, and we have, in the third premise, $\lnot t$. By modus tollens, we derive $\lnot p$.
Now, since we've derived $\lnot p$ from premises (3), (4), we also have $\lnot p \lor q$, by "addition" to $\lnot p$, (also called "or-introduction" which is shorthand for disjunction introduction).
And from $\lnot p \lor q$, together with the premise (1), we have $r$ by modus ponens.
Now, since we already deduced $\lnot p$, and we just deduced $r$, we can use "And-introduction (conjunction-introduction)" to get $\lnot p \land r$.
Given $\lnot p \land r$, and premise (5): $(\lnot p \land r) \to \lnot s$, we have, by modus ponens, $\lnot s$.
But given our premise (2), $s \lor \lnot q,$ together with $\lnot s$, we deduce $\lnot q$, as desired.
Best Answer
$\neg s$ premise $1$
$(p \wedge t) \to (r \lor s)$ premise $2$
$(p \land t) \to r$ based on $1$ and $2$ Disjunctive Syllogism
$q \to (u \land t)$ premise $4$
$q \to r$ (hypothetical syllogism) from $3$ and $4$
Is this answer correct? I am not sure if $(p \land t) == (u \land t)$. Because based on Hypothetical syllogism, $P \to Q$ and $Q \to R \implies P \to R$. In this cause I am taking $(p \land t) == (u \land t) == Q$