Extend the segment $AC$ beyond $C$ by a segment $CE$ congruent to the segment $CB$, so $C$ lies on the perpendicular bisector of the segment $BE$. Similarly, extend the segment $BD$ beyond $D$ by a segment $DF$ congruent to the segment $AD$, so $D$ lies on the perpendicular bisector of the segment $AF$. Since the triangles $ABC$ and $ABD$ have equal perimeters, the segments $AE$ and $BF$ are congruent.

Now let the perpendicular bisectors of the segments $AF$ and $BE$ meet at $Q$. Clearly, the segments $QA$ and $QF$, respectively $QE$ and $QB$, are congruent, and since so are the segments $AE$ and $BF$ by the preceding, the triangles $QAE$ and $QFB$ are congruent. Therefore, the angles $AQF$ and $BQE$ are congruent, and hence so are their halves; that is, the angles $AQD$ and $BQC$ are congruent, and consequently so are the angles $AQC$ and $BQD$.

We now show that, in fact, the points $P$ and $Q$ coincide, whence the conclusion. With reference again to the congruence of the triangles $QAE$ and $QFB$, the angles $QAE$ and $QFB$ are congruent, and since the latter is the reflection of the angle $QAD$ in the line $QD$, it follows that the angles $QAC = QAE$ and $QAD$ are congruent, so the line $AQ$ bisects the angle $CAD$. Similarly, the line $BQ$ bisects the angle $CBD$, and consequently the points $P$ and $Q$ coincide.