Figure 3

Without loss of generality let the points on the side $AC$ be in the order $A$, $K$, $L$, $C$ and on the side $BC$ in the order $C$, $M$, $N$, $B$ (see Fig. 3). Choose the point $P$ so that the quadrilateral $LCMP$ is a square. Then $|KL| = |LC| = |LP|$ and $|MN| = |CM| = |MP|$, i.e. $KLP$ and $PMN$ are isosceles right triangles, so $\angle KPL = \angle MPN = 45^\circ$. Since $\angle KPN = 45^\circ + 90^\circ + 45^\circ = 180^\circ$, the point $P$ lies inside the segment $KN$, whose all points except the endpoints are inside the triangle $ABC$.

To show that $P$ is the only point with the required properties, let $P'$ be an arbitrary point inside the triangle $ABC$ which satisfies $\angle KP'L = \angle MP'N = 45^\circ$. Since $P$ and $P'$ are on the same side of the line $KL$ and $\angle KPL = \angle KP'L$, the point $P'$ lies on the circumcircle of the triangle $KPL$; similarly it also lies on the circumcircle of the triangle $MPN$. Since $\angle KLP = \angle PMN = 90^\circ$, the segments $KP$ and $PN$ are the diameters of the circles. Since the diameters $KP$ and $PN$ lie on the same straight line $KN$, they have a common perpendicular at the point $P$ which is tangent to both circles at this point. Hence the point $P$ is the only common point of these circles, i.e. $P' = P$.