Proving the following proposition

Question

I am asked to prove the following proposition but, due to omissions or just plain ignorance I've been unable to arrive at the correct answer. Your input will be greatly appreciated.

Answer 1

Intuitively we can reason that no object $a$ can satisfy $p$ because if it did we'd also have that something is $q$ (by first premiss), which is ruled out (by the second premiss). So, turning that idea into a natural deduction proof:

Start a subproof with $p(a)$.

Then it follows $\exists x p(x)$, and hence by the first premiss $\exists y q(y)$.

But $\exists y q(y)$ and the second premiss lead to contradiction.

Since $p(a)$ leads to contradiction we can apply reductio, drop the supposition, and conclude $\neg p(a)$.

Universal quantifier introduction gives the target result.

How that goes in detail will depend on the system you are working with.