Two brown-haired people have two children, P and Q. P has blond hair. Q has brown hair. What is the probability that Q is heterozygous?
Assumptions: Blonde hair is Homozygous Recessive and that the traits are strictly Mendelian.

The parental generation must be both heterozygotes as at least one child is Blonde (bb). So your cross is Bb x Bb.

Your square is going to look like this:

           _B_            _b_

_B_         BB           Bb



_b_         bB           bb


So of the question is:


  What is the probability that Q is heterozygous?


The PhD is INCORRECT. They probably just made a simple Punnet Square in their head and forgot the caveat that you have to exclude "bb" because Q is not Blonde. The possible allele combinations for non-Blonde offspring from a Heterozygotic cross are only BB, bB, Bb. 

Therefor the probability that Q is Heterozygous is indeed 2/3 (bB, Bb).

It's not uncommon for PhD students - who are often deep into their own research - to gloss over details and possibly give incorrect answers because the information hasn't been relevant to them for years (most traits aren't Mendelian, strictly recessive or dominant, and rarely involve a single allele), so give them some slack unless they were being a jerk about it. =)