Why does the Complexity Class PostBQP makes proving PP greater than or equal to QMA easier? PP was proved greater than QMA in by Kitaev and Watrous 2000 -
Parallelization, Amplification, and Exponential Time Simulation of Quantum Interactive Proof Systems.

Later Aaronson proved that PostBQP was equal to PP in his PHD 2004.

It has been stated that the knowledge that PostBQP = PP makes proving PP is greater than or equal to QMA a lot easier. 

Is there a short explanation of this. Or if not could someone provide a reference detailing this fact?
I think the reason PostBQP contained in PP makes it easier to show QMA in PP is that it's easy to show QMA in PostBQP. (Note that we don't really need PostBQP contains PP, we only need the reverse containment, which is easier to show.)

QMA can be shown to be in PostBQP in exactly the same way that one would show MA is in PostBPP (also known as BPPpath). First use error amplification to reduce the error of the MA machine to something more than exponentially small in the size of the witness. Now if you use a random witness, in the yes case there will be some probability of it being accepted (although exponentially small), whereas in the no case the probability of acceptance will be even lower. In the quantum case, we choose the maximally mixed state as the witness and use strong error amplification for QMA. (Both of these ideas are explained in John Watrous' survey on quantum complexity theory.)

Now we need to use the power of post-selection to distinguish these two cases. This is not too hard, but it might sound complicated when I explain it. If you have a machine that accepts yes instances with probability $1/2^k$ (say) and no instances with probability $1/2^{k+4}$. Let our machine with the power of post-selection do the following: With probability $1/2^{k+2}$ just set the post-selection bit (let's call this the PSB) to 1 and the output bit to 0 and end the computation. Otherwise just run the previous machine and set the PSB and the answer bit equal to the output of the computation.

Now in the yes case, with probability $1/2^{k+2}$ the machine with have the PSB equal to 1 and answer bit equal to 0, and with probability at least $1/2^k$ it will have the PSB equal to 1 and answer bit equal to 1. So conditioned on the PSB being 1, there's a much higher chance that it accepts than rejects. Similarly, in the no case, conditioned on the PSB being set to 1 there will be a higher chance of rejecting than accepting.

You might have to tweak the constants a bit to make this idea work exactly, but I think it contains all the ideas needed to make the proof go through.