If $P=BPP$, then Is it correct that $IP=NP$?
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...