Wlog $PA=1$. It's well known that $PA^2+PC^2=PB^2+PD^2$ (for example, take a point $D'$ not inside the square such that the triangles $CPD$ and $BD'A$ are congruent, show that $PD'\perp AB$ and finally $PA^2+BD'^2=PB^2+AD'^2$, or $PA^2+PC^2=PB^2+PD^2$). This gives us $PD=\sqrt{6}$. Take a point $E$ not inside the square such that the triangle $APB$ is congruent to $AED$, so $AE=AP=1$ and $\angle{PAD}+\angle{BAP}=\angle{PAD}+\angle{DAE}=90^{\circ}$ so $AP\perp AE$, $PE=\sqrt{2}$ and $\angle{PED}=\angle{APB}-45^{\circ}$. Finally in triangle $PED$, we have $DE^2+EP^2=6=PD^2$, so $\angle{PED}=90^{\circ}$ gives $\angle{BPA}=135^{\circ}$.

Wlog, the sidelength of square is $1$. Let us rotate the square $90^\circ$ degrees from $B$ anticlockwise. Call this new rotation of $P$ as $P'$. Note that $BPP'$ is a isosceles right triangle. Thus, $PP'=2\sqrt{2}$. Now, by Pythagorean theorem, $PC^2=P'C^2+PP^2$, hence $\angle PP'C=90^\circ$. We conclude that $\angle BPA=\angle BP'C=45^\circ+90^\circ=\boxed{135^\circ}$.

If one does not find that clever solution, there is also a straightforward solution by complex numbers: Let $P=0, A=1, B=z$ so that $\vert z\vert=2$ i.e. $\overline{z}=\frac{4}{z}$. Now $C=z+i(z-1)$ and hence the condition becomes $\vert z+i(z-1)\vert=3$ i.e. \[9=(z+i(z-1))\left(\frac{4}{z}-i\left(\frac{4}{z}-1\right)\right)\]and hence \[9z=(z+i(z-1))(4+i(z-4))=(i-1)z^2+9z+4i(i-1)\]and hence $z^2=-4i$ from where we can immediately read off the angle.

Interestingly, of course both solution work if we replace $1:2:3$ by any $a:b:c$ with $a^2+2b^2=c^2$ which shows that the special choice of $1:2:3$ is really a bit of a red herring.