You're told the ka for benzoic acid, and then you're told that you have a bunch of benzoate in solution, the conjugate base of benzoic acid. Since you have the conjugate base, you want to use the kb of that base for the reaction:
benzoate + H20 ---> Benzoic acid + OH-
so kb = x^2/0.5M where x = the concentration of OH- ions and benzoic acid.
so you do the math and find the concentration of OH- slash benzoic acid, and from there the easiest thing is take the -log of OH- to get the pOH and then subtract that from 14 to get the pH. You can get the concentration of hydronum ions by dividing kw by hydroxide concentration or take 10^-pH.
As TS implied in his answers, so little benzoate picks up protons that the amount that it's reduced by is negligible. This should be the case with any MCAT question because no calculators- you need to be able to solve questions like this without dicking around with the quadratic equation. That goes for common ion effect problems too.