There are a couple of ways to solve this. Perhaps simplest is to realize the figure is symmetrical about the line AP. Then angle BAP = angle ABP = 20°. Half of angle x° will be the sum of BAP and ABP, so is 40°. Since that is half of x, the full value of x is 80.
Another way to solve this is to consider that the sum of angles inside the quadrilateral ABPC is 360°. You know that angle BAC = x/2, so you have
.. x°/2 +20° +(360°-x) +20° = 360°
.. -x°/2 = -40° . . . . . collect terms, subtract 400°
.. x = 80