If P = BQP, does this imply that PSPACE (= IP) = AM? Recently, Watrous et al proved that QIP(3) = PSPACE a remarkable result. This was a surprising result to myself to say the least and it set me off thinking...

I wondered what if Quantum Computers could be efficiently simulated by Classical Computers. Could this be SIMPLY related to the Divide between IP and AM? What I mean is that IP is characterized by Polynomial number of rounds of  classical interaction, while AM has 2 rounds of classical interaction. Could simulating a Quantum Computing reduced the amount of interaction for IP from polynomial to a constant value?
Great question!  Short answer: no implication like
$$
\mathsf{P} = \mathsf{BQP}   \Rightarrow  \mathsf{IP} = \mathsf{AM}
$$
is known; but that doesn't mean it's not worth trying to prove... 

I would say, though, that finding such an implication seems unlikely.  I think the message of much quantum complexity theory is that, while quantum computers are not an all-purpose panacea for solving hard problems, they can be much more powerful than classical computers in certain specific circumstances.  

For example, in query complexity, quantum algorithms can efficiently solve certain problems classical ones provably cannot, when the input is promised to obey some nice global structure.  E.g., Shor's algorithm is based on an algorithm to quickly find the unknown period of a function promised to be periodic.  On the other hand, quantum query algorithms are not too much stronger than classical ones for solving problems in which there is no special structure assumed on the input.  (See Buhrman and de Wolf's survey on query complexity for this last point.)

Similarly, I think the results $\mathsf{QIP}(3) = \mathsf{QIP} = \mathsf{IP}$ tell us, not that interaction is unexpectedly weak (even if $\mathsf{P} = \mathsf{BQP}$), but that quantum computation is unexpectedly strong, specifically in the context of interaction with computationally-unbounded provers.