EDIT: didn't do angles correctly so have to redo it.

EDIT2:

I did not calculate or make notes but I found the path to the solution. It is just a matter of calculating angle after angle after angle.

But you can... it's an awful lot of angles before you get to it though

I am strongly of the opinion that this cannot be solved simply by filling in angle values using geometric rules. Although the answer does come out to an integer, adjusting one of the starting angles slightly to a different integer results in the answer coming out as irrational, which would thus not be obtainable by application of simple geometric rules. So I believe trigonometry is required. Which I applied to get my solution, which I will withhold for now.



EDIT: Here's a modified picture with some suggestions which hint to how I solved the problem with trigonometry and algebra.



EDIT 2: And here's the worked out solution. Each step is nested in a spoiler, so only peek as much as you need to!

Hey, I was passing by and I just saw this topic.
Runner's solution works, but using analysis for such a problem is sad.

I found this nice solution in 5 minutes :

- Clearly PCA=PAD=20°
- Let O be the circle that passes through P, C, A
- Since C and A are on O and PCA=PAD, DA is tangent to O in A (corollary of inscribed angles theorem)
- DA and AB are perpendicular so AB passes through the center of O
- Since C and A are on O and PCD=PAC, DC is tangent to O in C
- DC and CB are perpendicular so CB passes through the center of O
- Hence the center of O is the intersection of AB and CB, B
- Since B is the center of O and P, A belong to O, PBA=2*PCA=40°

I hope you can understand despite my poor English

Nice one Chockosta! Here a little drawing showing your solution:
   
1. Assume B is the center of the drawn circle (ABCD is a square!).
2. BP and BA are both radii of the circle (have equal lengths) => ABP triangle is isosceles!
3. Meaning, angles BPA=PAB=70º => ABP = 180 - 140 = 40º!

Thanks for those drawings, but they only show the easy part of the proof.

The interesting part is that B is the center of the drawn circle, which is true if and only if PAC=PCD.
Here are some drawings :


The inscribed angle theorem says that the blue angle remains the same when you move B.
(This is a well-known and useful result)
The corollary of this theorem says that it is the same as the angle with the tangent in C.
(To understand this, keep in mind that the blue angle doesn't change when B is getting close to C)


In this drawing you can notice that the purple angles are acting the same way as mentionned before.
Therefore, you can say that AD is tangent to the circle in A.
Moreover, you can do the same with the blue angles. So CD is tangent to the circle in C.


Since ABCD is a square, CB is perpendicular to CD in C. But CD is tangent to the circle in C.
Therefore CB is a radius of the circle.
You can do the same and find that AB is also a radius.
The two radii meet in B, so B is the center of the circle.

The conclusion is now quite obvious, see SpiroH's drawings
I personnally did it with the inscribed angle theorem :
You can apply this to PAC and PBC and find out that PBC=2*PAC, so PBC=40°