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, $\angle PBP'=90^{\circ}$ and
$BP=BP'=2$. By Pythagoras, $PP'= \sqrt 8$. Now consider triangle
$AP'P$. The side lengths are $3$, $\sqrt 8$, and $1$ which
satisfies Pythagoras again: $3^2 - 1^2 = (\sqrt 8)^2$.
So this triangle has $90$ degrees at $\angle P'PA$. Since we
know that $\angle BPP'=45^{\circ}$ therefore $\angle BPA$ is $135$
degrees.
This alternative method comes from Yatir from Maccabim-Reut High
School, Israel.
$AB=BC=CD=AD=x$, $BP=2$, $CP=3$, $AP=1$. Name the angles this way:
$\angle ABP = c$, $\angle CBP = 90-c$, $\angle APB = a$, $\angle
CPB = b$.
Using the Cosine Law we find $\cos b$ and then eliminate $b$: $$x^2
= 9 + 4 - 12 \cos b$$ so $$\cos^2 b = [(13-x^2)/12]^2.$$
Using the identity $\sin^2 b + \cos^2 b = 1$ we get $$(x^2 - \sin
^2 a)/9 + [(13 - x^2)^2]/144 = 1 \quad (1).$$
Finally we eliminate $x$. By the Cosine Law $x^2 = 4 + 1 - 4\cos a$
so
$$x^2 = 5 - 4 \cos a \quad (2).$$
Combine (1) and (2) and simplify and we get: $$2 \cos^2 a /9 + 8 /9
= 1$$ hence $\cos^2 a = 1/2.$
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.