Let $AC = CB = 3a$. Clearly the set of points $K$ such that $\angle KPL = 45^\circ$ is the circumfrence of a circle passing through $K$ and $L$ such that the arc $KL$ is $90^\circ$. The analogous result is true for $\angle MPN$ as well. We proceed with coordinates.
The center of the circle passing through $K$ and $L$ is $\left( \frac{a}{2}, \frac{3a}{2} \right)$ and the center of the other circle is $\left( \frac{3a}{2}, \frac{a}{2} \right)$. Both circles have radius $\frac{a\sqrt{2}}{2}$ as well. Thus the equations of the cicles going through $K$ and $L$ alongside the circle going through $M$ and $N$ are:
$$\left( x-\frac{a}{2} \right)^2 + \left( y-\frac{3a}{2}\right)^2 = \frac{a^2}{2}$$and $$\left( x-\frac{3a}{2} \right)^2 + \left( y-\frac{a}{2}\right)^2 = \frac{a^2}{2}$$respectively. It can be shown through equating both LHS to get $2ax-2ay=0$ that $x=y$. Thus, there is only one solution, more precisely $(a, a)$ and we are done. $\square$