How does the Pusey-Barret-Rudolph (PBR) theorem not just disprove hidden variables? In Quantum Mechanics, two different wavefunctions can have a non-zero probability of finding a particle at a position $x$.
According to hidden variable theories, if a particle is found at $x$, it was there before the measurement. $x$ was the hidden variable.
According to PBR theorem, a single hidden variable $x$ can not be consistent with multiple wavefunctions. The assignment of hidden variables to wavefunctions cannot be many-to-one. This contradicts the first paragraph of the post
Since the first paragraph is true, doesn't this mean that hidden variables,  including Bohmian mechanics, are disproved?
 A: It does not "disprove" hidden variables, but rather says that, under a suitable formulation and assumptions about what form the hidden variable theory takes, that the hidden variables must be at least as complex as the quantum Hilbert space. That is to say, there is a(n almost) distinct class of hidden-variable assignments for each and every of the uncountably many, $\beth_1$, possible vectors in the Hilbert space or, if you like, that the hidden variables can be described or labeled in such a way that at least one of those variables must be equivalent to the quantum vector itself.
The idea behind the PBR theorem is this. Consider a coin in classical probability theory - not one we're going to flip, but one we're going to have some other device give us in either the heads or tails conditions, randomly, according to some settings. If that device is set up to choose heads with some probability $p$, then a person ignorant of exactly what the coin has been flipped to, but knows the setting on the device, must say that it is heads only with probability $p$. Moreover, by setting $p$ suitably on the device, we can convey to someone else a "heads" with any real-number probability $p$ even though the coin itself has only two states.
Then, under suitable combining assumptions for how multiple coins' states combine, if we have a preparation system for a quantum "coin", the theorem tells us we cannot treat the whole continuum of quantum probabilities about it as simply being due to that it has a few states and the preparation system generates those probabilities by mixing them together. Instead, it is like the coin really has not only infinitely many states, but so many there is at least one for each analogue of the real number $p$ in the classical coin case (for a qubit, this means a point on the Bloch sphere) or to say it another way, it is as though the coin carried the whole probability setting that was on the preparer. Or else, either the combining assumptions are wrong (though I've also heard that isn't "strong enough" to defeat the theorem), or, in a result reminiscent of Bell's Theorem but for the preparer, multiple preparers must somehow "spookily" correlate their preparations non-locally.
A: In the post, I assumed that $x$ completely specifies the hidden variable state. But that's not true because the two states can have the same $x$ but differ in $p$. Hence the combined states $(x, p) $ are different and there's no contradiction with PBR.
