Another way to prove it

Let C bet the focus, and C' be another focus, O be the original, and PQ cross the x axis at R. Prove that the triangle PP'C is similar to triangle CPR. Actually, angle P'PC = angle PCR, and P'P/Pc = a/c, by the definition of the directrix.

Need only prove that PC/CR = a/c as well. (I did not do the calculation, but it should be if the conclusion is correct).

Once it is done, we have angle RPC (i.e. angle QPC) = angle CP'P = angle P'CX = angle OCQ = angle CC'Q.