If $P=BPP$, then Is it correct that $IP=NP$? This is my first question in this site. I ask this question since I got no comment and no answer for one year and two months in cs.stackexchange and it was automatically deleted by the system. So, this is not a repeated question, my question is the following:

I'm now studying Interactive Proof System in Goldreich's textbook and here. I have the following question:

Suppose P=BPP, then this show that randomness doesn't give us power over deterministic model (moreover, this would show that we can derandomize every randomness algorithm into deterministic model). Now, In interactive proof system, we know that if Verifier is deterministic, then Prover can guess all questions that Verifier would ask. So, interactive proof system is useless without randomness (see Claim 1 in the above lecture note). Moreover, interactive proof system such that Verifier is not using randomness would be equivalent to class NP. Now, does this mean if P=BPP, then IP=NP? Why I say something like that? Well, since Prover has unbounded computation power, then Prover can simulate randomness by derandomness.

Thank you!
This is not known, but as domotorp stated, it is believed not to be the case. First, note that $\mathsf{P} = \mathsf{BPP}$ doesn't say that randomness isn't useful in any context, just in the context of poly-time decision problems. For example, just assuming $\mathsf{P} = \mathsf{BPP}$ is already not known to imply that $\mathsf{AM} = \mathsf{NP}$ (and the latter is even weaker than $\mathsf{IP} = \mathsf{NP}$). To give you a little intuition, the point is this: $\mathsf{P} = \mathsf{BPP}$ says that when deciding a deterministic thing, randomness doesn't help (up to poly factors). But in interactive proofs, the randomness isn't used to decide a deterministic thing, the randomness is used to "fool" the prover. 

As a related (but admittedly different) example note that even if $\mathsf{P} = \mathsf{BPP}$, that doesn't obviate the need for randomness in distributed computing. For example, selecting unique IDs for processors in a distributed system can be done using randomness with high probability (everyone just picks a random ID from a large enough range), and even if $\mathsf{P} = \mathsf{BPP}$ there seems to be essentially no way to remove this use of randomness. (Same goes for one-time pads in crypto.)

I suppose some more formal evidence would be an oracle relative to which $\mathsf{P} = \mathsf{BPP}$ and yet $\mathsf{IP} \neq \mathsf{NP}$ - shouldn't be too hard to construct...