Consider the following complexity class $IP^\star$, a variant of $IP$. A language $L$ is in $IP^\star$ if there's a proof system $(P,V)$ s.t. $V$ is a verifier runs for a polynomial time and:
$$x\in L \implies Pr[(P,V)(x) = 1] \ge 3/4 \\ x\not\in L \implies > Pr[(P,V)(x) = 1] = 0 \\ $$
Prove that $NP = IP^\star$.
Note: in this definition as far as I understand, we have a single $P$ (for both cases).
Now, one side is easy ($NP\subseteq IP^\star$), since by definition $NP$ language has a perfect polynomial-verifier - so we're satisfying the demands.
The other side is the tricky one. Consider a language $L\in IP^\star$. We want to show that $L\in NP$. Obviously we shall utilize $V$ somehow. The certificate could be the "conversation" between $V$ and $P$. Then our $NP$ TM could simulate the conversation between $P$ to $V$.
Yet, that's alone isn't good enough - we have a $1/4$ probability to make a mistake if $x\in L$.
Amplifying the probability isn't good enough too. We need perfect soundness.
So maybe we shall use the non-determinism of our machine somehow?