$$(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$$ is equivalent to $$(\sim p)$$ when $$r$$ is

Given,

$$(p\wedge r)\Leftrightarrow(p\wedge(\sim q))\equiv(\sim p)$$

Now,

$$A\Leftrightarrow B\equiv(A\wedge B)\vee((\sim A)\wedge(\sim B))$$

Let

$$A=(p\wedge r),\qquad B=(p\wedge(\sim q))$$

Then,

$$(p\wedge r)\Leftrightarrow(p\wedge(\sim q))$$

is true when both have same truth values.

Now consider cases.

Case 1: $$p=\text{FALSE}$$

Then,

$$p\wedge r=\text{FALSE}$$

and

$$p\wedge(\sim q)=\text{FALSE}$$

Hence,

$$\text{FALSE}\Leftrightarrow\text{FALSE}=\text{TRUE}$$

Also,

$$(\sim p)=\text{TRUE}$$

Thus equivalence holds.

Case 2: $$p=\text{TRUE}$$

Then expression becomes

$$r\Leftrightarrow(\sim q)$$

But RHS is

$$(\sim p)=\text{FALSE}$$

Hence,

$$r\Leftrightarrow(\sim q)=\text{FALSE}$$

This happens when

$$r\ne(\sim q)$$

i.e.,

$$r=q$$

Therefore,

$$r\equiv q$$

Hence, the required answer is

$$\boxed{q}$$.