∆AQB is isosceles, so ∠QAB = ∠QBA.
The sum of vertex angles of ∆ABC is 180° so you have
... ∠QAB + ∠QCB + (∠QBC + ∠QBA) = 180°
Substituting what we know ...
... ∠QAB + 52° + (28° + ∠QAB) + 180°
... 2∠QAB = 100°
... ∠QAB = 50°
Because PQ is parallel to BC, ∠APQ = ∠ABC = 50° + 28°
... ∠APQ = 78°
