This beautifully neat solution was sent in by Johnny Chen.

If we rotate the figure by 90 degrees about B (clockwise). This should result in a new square A ' BC ' D ' with a point P ' inside the new square such that, ∠PBP ' =90^o and BP=BP ' =2. By Pythagoras, PP ' = √8. Now consider triangle AP ' P. The side lengths are 3, √8, and 1which satisfies Pythagoras again: 3² − 1² = (√8)².

So this triangle has 90 degrees at ∠P ' PA. Since we know that ∠BPP ' =45^o therefore ∠BPA is 135 degrees.

This alternative method comes from Yatir Halevi, age 18 from Maccabim-Reut High School, Israel.

AB=BC=CD=AD=x, BP=2, CP=3, AP=1. Name the angles this way: ∠ABP = c, ∠CBP = 90−c, ∠APB = a, ∠CPB = b.

We need to find angle a. By the Sine Law:

x sinb =  3 sin(90−c) =  3 cosc

and

x sina =  1 sinc

Hence sinc = sina /x, cosc = 3sinb /x and we use the identity sin² c + cos² c = 1 to eliminate c. So

sin2 (a/x2) + 9sin2 (b / x2) = 1 sin2b = (x2 − sin2 a)/9.

Using the Cosine Law we find cosb and then eliminate b:

x2 = 9 + 4 − 12 cosb

so

cos2 b = [(13−x2)/12]2.

Using the identity sin² b + cos² b = 1 we get

(x2 − sin2 a)/9 + [(13 − x2)2]/144 = 1     (1).

Finally we eliminate x. By the Cosine Law x² = 4 + 1 − 4cosa so

x2 = 5 − 4 cosa     (2).

Combine (1) and (2) and simplify and we get:

2 cos2 a /9 + 8 /9 = 1

hence cos² a = 1/2.

This yields solutions: a=45 or a=135 degrees.

Now this angle cannot be 45 degrees because angle c is smaller (AP is the shortest side of triangle APB) and the third angle of the triangle is less than 90 degrees. Hence angle a is 135 degrees.