Why $1,2,3$ ?

Rotate the square through 90° around point B. Let P' be the image of P, A' that of A, and D' of D. Connect P and P'.

NOTE : APC and AP'A' aren't straight lines although they might look so.

In ΔBPP', $BP = BP'$ and $\angle P'BP=90^\circ$ . The triangle is also isosceles so that $\angle BPP' = 45^\circ$ . From the Pythagorean Theorem, $PP^2 = 8.$

In ΔAPP', $PP^2 + AP^2 = AP^2(8 + 1 = 9)$ . Therefore, by the converse of the Pythagorean Theorem, the triangle is right and $\angle P'PA = 90^\circ$ .

Summing up, $\angle APB = \angle BPP' + \angle P'PA = 45 ^ \circ+90 ^ \circ=135^\circ$ .

Let the side length of square $ABCD$ be $2a$ and the center of the square be the origin of the $xy$ -plane such that the coordinates of the vertices of the square be $A(-a,a)$ , $B(a,a)$ , $C(a,-a)$ , and $D(-a,-a)$ , and $P(x,y)$ . By Pythagorean theorem :

$\begin{cases} AP: & (x+a)^2 + (y-a)^2 = 1^2 & ...(1) \\ BP: & (x-a)^2 + (y-a)^2 = 2^2 & ...(2) \\ CP: & (x-a)^2 + (y+a)^2 = 3^2 & ...(3) \end{cases}$

$\begin{aligned} (1)-(2): \quad 4ax & = - 3 & \implies x = - \frac 3{4a} \end{aligned}$

$\begin{aligned} (3)-(2): \quad 4ay & = 5 & \implies x = \frac 5{4a} \end{aligned}$

$\begin{aligned} (1): \quad \left(-\frac 3{4a}+a\right)^2 + \left(\frac 5{4a}-a\right)^2 & = 1 \\ 2a^2 - 5 + \frac {17}{8a^2} & = 0 \\ 16a^4 - 40a^2 + 17 & = 0 \\ \implies a^2 & = \frac {5 \pm 2\sqrt 2}4 \end{aligned}$

By cosine rule :

$\begin{aligned} AB^2 & = AP^2 + BP^2 - 2AP\times BP \cos \color{#3D99F6} \angle APB & \small \color{#3D99F6} \text{Let }\angle APB = \theta \\ 4a^2 & = 1 + 4 - 4 \cos \theta \\ \cos \theta & = \frac {5-\color{#3D99F6}4a^2}4 & \small \color{#3D99F6} \text{Note that }a^2 = \frac {5 \pm 2\sqrt 2}4 \\ & = \frac {5-\color{#3D99F6}5 \mp 2\sqrt 2}4 \\ & = - \frac 1{\sqrt 2} & \small \color{#3D99F6} \text{Since } \theta > 90^\circ \\ \implies \theta & = \boxed{135}^\circ \end{aligned}$