This beautifully neat solution was sent in by Johnny Chen.

So this triangle has 90 degrees at $\angle P\text{'}\mathrm{PA}$. Since we know that $\angle \mathrm{BPP}\text{'}={45}^o$ therefore $\angle \mathrm{BPA}$ is 135 degrees.

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

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

${\displaystyle \frac{x}{\sin b}=\frac{3}{\sin (90-c)}=\frac{3}{\cos c}}$

and

${\displaystyle \frac{x}{\sin a}=\frac{1}{\sin c}}$

Hence $\sin c=\sin a/x$, $\cos c=3\sin b/x$ and we use the identity ${\sin }^{2}c+{\cos }^{2}c=1$ to eliminate $c$. So

${\displaystyle \begin{array}{cc}\multicolumn{1}{c}{{\sin }^2(a/x^2)+9{\sin }^2(b/x^2)}& =1\\ \multicolumn{1}{c}{{\sin }^{2}b}& =(x^2-{\sin }^{2}a)/9.\end{array}}$

Using the Cosine Law we find $\cos b$ and then eliminate $b$:

${\displaystyle x^2=9+4-12\cos b}$

so

${\displaystyle {\cos }^{2}b=[(13-x^2)/12{]}^2.}$

Using the identity ${\sin }^{2}b+{\cos }^{2}b=1$ we get

${\displaystyle (x^2-{\sin }^{2}a)/9+[(13-x^2{)}^2]/144=1\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1).}$

Finally we eliminate $x$. By the Cosine Law $x^2=4+1-4\cos a$ so

${\displaystyle x^2=5-4\cos a\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(2).}$

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

${\displaystyle 2{\cos }^{2}a/9+8/9=1}$

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 ( $\mathrm{AP}$ is the shortest side of triangle $\mathrm{APB}$) and the third angle of the triangle is less than 90 degrees. Hence angle $a$ is 135 degrees.