Q.Without using truth table show that $\sim (p \lor q) \lor (\sim p \land q) \equiv \sim p$

Apply De Morgan's law and then the distributive law.

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

Apply De Morgan's law to the first term: $\sim(p\lor q)\equiv\sim p\land\sim q$.

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

Apply the distributive law $A\land B\lor A\land C\equiv A\land(B\lor C)$ with $A=\sim p$, $B=\sim q$, $C=q$:

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

Since $\sim q\lor q$ is a tautology (always true, denote $T$):

$\equiv \sim p\land T \equiv \sim p$

…