$$p \to \left( { \sim p \vee q} \right)$$    has truth value $$F.$$
It means $$p \to \left( { \sim p \vee q} \right)$$    is false.
It means $$p$$ is true and $${ \sim p \vee q}$$   is false.
$$ \Rightarrow p$$  is true and both $$\sim p$$  and $$q$$ are false.
$$ \Rightarrow p$$  is true and $$q$$ is false.

Suman is brilliant and dishonest if and only if Suman is rich is expressed as
$$Q \leftrightarrow \left( {P \wedge \sim R} \right)$$
Negation of it will be $$ \sim \left( {Q \leftrightarrow \left( {P \wedge \sim R} \right)} \right)$$

$$p :$$ It rains, $$q :$$ I shall go to school
Thus, we have $$p \Rightarrow q$$
Its negation is $$ \sim \left( {p \Rightarrow q} \right){\text{i}}{\text{.e}}{\text{., }}p \wedge \sim q$$
i.e., It is rains and I shall not go to school.

$$p$$	$$q$$	$$p \to q$$	$$ \sim p$$	$$ \sim p \vee q$$	$$\left( {p \to q} \right) \leftrightarrow \left( { \sim p \vee q} \right)$$
T	T	T	F	T	T
T	F	F	F	F	T
F	T	T	T	T	T
F	F	T	T	T	T

$$p$$	$$q$$	$$ \sim q$$	$$p \leftrightarrow \, \sim q$$	$$ \sim \left( {p \leftrightarrow \, \sim q} \right)$$
F	F	T	F	T
F	T	F	T	F
T	F	T	T	F
T	T	F	F	T

Clearly equivalent to $$p \leftrightarrow q$$

The truth table for the logical statements, involved in statement - 1, is as follows :

$$p$$	$$q$$	$$ \sim q$$	$$p \leftrightarrow \, \sim q$$	$$ \sim \left( {p \leftrightarrow \, \sim q} \right)$$	$$p \leftrightarrow q$$
T	T	F	F	T	T
T	F	T	T	F	F
F	T	F	T	F	F
F	F	T	F	T	T

We observe the columns for $$ \sim \left( {p \leftrightarrow \sim q} \right)$$   and $${p \leftrightarrow q}$$   are identical, therefore
$$ \sim \left( {p \leftrightarrow \sim q} \right)$$   is equivalent to $${p \leftrightarrow q}$$
But $$ \sim \left( {p \leftrightarrow \sim q} \right)$$   is not a tautology as all entries in its column are not $$T.$$

$$\eqalign{ & \sim \left( {p \vee \left( { \sim p \wedge q} \right)} \right) \cr & = \sim \left( {\left( {p \vee p} \right) \wedge \left( {p \vee q} \right)} \right) \cr & = \sim \left( {t \wedge \left( {p \vee q} \right)} \right) \cr & = \sim \left( {p \vee q} \right) \cr & = \sim p \wedge \sim q \cr} $$

$$p :$$  A number is a prime
$$q :$$  It is odd.
We have, $$p \Rightarrow q$$
The inverse of $$p \Rightarrow q{\text{ is }}\sim p \Rightarrow \sim q$$
i.e., if a number is not a prime then it is not odd.

Check each option
$$\eqalign{ & \left( 1 \right)\,\,\left( {p \vee q} \right) \wedge \left( { \sim p \wedge q} \right) = \left( { \sim p \wedge q} \right) \cr & \left( 2 \right)\,\,\left( {p \vee q} \right) \wedge \left( { \sim p \vee q} \right) = \left( {p \wedge \sim p} \right) \vee q \cr & = F \vee q = q \cr & \left( 3 \right)\,\,\left( {p \wedge q} \right) \wedge \left( { \sim p \vee q} \right) = \left( {p \wedge q \wedge \sim p} \right) \vee \left( {p \wedge q} \right) \wedge q \cr & = F \vee \left( {p \wedge q} \right) = p \wedge q \cr & \left( 4 \right)\,\,\left( {p \wedge q} \right) \wedge \left( { \sim p \wedge q} \right) = \left( {p \wedge \sim p} \right) \wedge q \cr & = F \sim q = F \cr} $$

Let us make the truth table for the given statements, as follows :

$$p$$	$$q$$	$${p \vee q}$$	$$q \to p$$	$$p \to \left( {q \to p} \right)$$	$$p \to \left( {p \vee q} \right)$$
T	T	T	T	T	T
T	F	T	T	T	T
F	T	T	F	T	T
F	F	F	T	T	T

From table we observe
$$p \to \left( {q \to p} \right)$$   is equivalent to $$p \to \left( {p \vee q} \right)$$