Answer:
Option A
Explanation:
Key Idea De- Morgan's and distributive law
We have $\sim (p \vee q) \vee(\sim p\wedge q)$
= $(\sim p\wedge\sim q)\vee (\sim p \wedge q)$ [ $\therefore$ By De-Morgan's law]
$\sim (p\vee q)=(\sim p\wedge \sim q)$
≡$\sim p \wedge (\sim q\vee q) $ [ By distributive law]
≡ $\sim p \wedge t $ [ $\sim q \vee q=t $]
≡ $\sim p$