Given expression: $(p \wedge \neg q) \wedge (p \wedge r) \rightarrow \neg p \vee q$
Let’s assume, $\alpha$ = $(p \wedge \neg q) \wedge (p \wedge r)$; $\beta$ = $\neg p \vee q$
Question is saying given expression is FALSE.
Implication is only FALSE for one condition $T \rightarrow F$. So, here $\alpha$ must be TRUE & $\beta$ must be FALSE.
So, $\beta$ = $\neg p \vee q$ = $F$
($\beta$ will be FALSE only if $\neg p$ and $q$ both are false.) [OR is only FALSE when both are false]
So, $\neg p$ = $F$; $q$ = $F$
Now we got, $p$ = $T$ and $q$ = $F$
Now $\alpha$ must be TRUE.
So, $\alpha$ = $(p \wedge \neg q) \wedge (p \wedge r)$ = $T$
We can clearly see that it will be TRUE only if $(p \wedge \neg q)$ and $(p \wedge r)$ both are TRUE.
$(p \wedge \neg q) = (T \wedge \neg F) = (T \wedge T) = T$
$(p \wedge r) = (T \wedge r) = T$ ← This will be TRUE only if r is TRUE. [AND is TRUE only in one condition $T \wedge T = T$]
So, $r$ = $T$.
Finally we got the truth values of $p, q, r$ as $T, F, T$ respectively.
Option (B) is correct.