How do we prove NP in BPP implies PH in BPP? Let me just do NPNP ⊆ BPP, the rest is an easy induction. We use the following inclusions.
The first and third "⊆" are by assumption, the "=" is by simulation. To get NPBPP ⊆ BPPNP we first make the error of the BPP algorithm so small that a randomly chosen string will give, with high probability, the correct answer for every query made on every computation path of the NP algorithm. We can then safely pick a random string r first and then make an NP query that simulates the NP algorithm which will use r to simulate the BPP queries.
Now suppose we wanted to show NP in BQP implies PH in BQP. We just need to show NPBQP ⊆ BQPNP, the rest works as before. Like the BPP case we can make the error of the BQP algorithm very small, but we have no quantum string that we can choose ahead of time. In probabilistic algorithms we can pull out randomness and leave a deterministic algorithm with a random input but we don't know any way to pull the quantumness, even with quantum bits, out of a quantum algorithm.
Whether NPBQP ⊆ BQPNP remains an open question. Showing NPBQP ⊆ BQPNP or finding a relativized world where NPBQP is not contained in BQPNP would give us a better understanding of the mysterious nature of quantum computation.