Let $R_A$ be the exsimilicenter of $\omega_B,\omega_C$. Note $R_A \in BC$. Define $R_B,R_C$ similarly. The Monge D'Alembert theorem gives that $\overline{R_AR_BR_C}$, $\overline{R_AP_BP_C}$, $\overline{R_BP_CP_A}$, $\overline{R_CP_AP_B}$ are collinear. But then $\triangle P_AP_BP_C$ and $\triangle ABC$ are perspective from line $\overline{R_AR_BR_C}$, so by Desargue's theorem they are perspective from a point and the result follows.

We use the notation $E(\Gamma_1, \Gamma_2)$ for exsimilicentre $\Gamma_1, \Gamma_2$ respectively. Define $E_A = E(\omega_B, \omega_C)$ , $E_B = E(\omega_A, \omega_C)$, $E_C = E(\omega_A, \omega_B)$. By Monge's theorem on $(\omega_A, \omega_B, \omega_C)$, we get these tuples of points are colinear: $(E_A, E_B, E_C)$, $(P_A, P_B, E_C)$, $(E_A, P_B, P_C)$, $(P_A, E_B, P_C)$. Then the triangles $(\Delta ABC, \Delta P_AP_BP_C)$ are perspecive axially. By Desargue's, they're perspective centrally as desired.

Nice problem math154 wrote: Let $ABC$ be a triangle. Draw circles $\omega_A$, $\omega_B$, and $\omega_C$ such that $\omega_A$ is tangent to $AB$ and $AC$, and $\omega_B$ and $\omega_C$ are defined similarly. Let $P_A$ be the insimilicenter of $\omega_B$ and $\omega_C$. Define $P_B$ and $P_C$ similarly. Prove that $AP_A$, $BP_B$, and $CP_C$ are concurrent. Tom Lu. Lemma: $\triangle O_AO_BO_C$ and $\triangle P_AP_BP_C$ are perspective , where $O_A$ is the center of $\omega_A$ and similarly others. Proof. Let $P_BP_C \cap O_BO_C = A_1$.Define $B_1,C_1$ analogously. Then as $P_B, P_C$ are the in-similicenters of of $\{\omega_A , \omega_C\}$ and $\{\omega_A, \omega_B\}$ respectively, it follows that $A_1$ is the ex-similicenter of $\{\omega_B, \omega_C\}$. Similarly $B_1$ and $C_1$ are the ex-similicenters of $\{\omega_A,\omega_C\}$ and $\{\omega_A , \omega_B\}$ respectively $\overset{\text{Monge's}}{\implies} A_1 - B_1 - C_1$. Hence by Desargue's theorem the lemma follows. $ $ Now since $BC$ is the common tangent to $\{\omega_B , \omega_C\} \implies A_1 \in BC$. Consequently, $B_1 \in AC$ and $C_1 \in AB$. Another application of Desargue's theorem on $\triangle ABC$ and $\triangle P_AP_BP_C$ finishes the problem $\blacksquare$

ELMOSL 2011 G3 wrote: Let $ABC$ be a triangle. Draw circles $\omega_A$, $\omega_B$, and $\omega_C$ such that $\omega_A$ is tangent to $AB$ and $AC$, and $\omega_B$ and $\omega_C$ are defined similarly. Let $P_A$ be the insimilicenter of $\omega_B$ and $\omega_C$. Define $P_B$ and $P_C$ similarly. Prove that $AP_A$, $BP_B$, and $CP_C$ are concurrent. Tom Lu. This Theorem Triviliases this Problem. Theorem:- Two of the Insimilicenters determined by three distinct circles all lying in the same plane are collinear with the Exsimilicenter of the last pair of circles. So Define $\{X_A,X_B,X_C\}$ as the Exsimilicenters of $\{(\omega_B,\omega_C),(\omega_A,\omega_C),(\omega_A,\omega_B)\}$. So, by Theorem we get that $\{\overline{P_A-P_B-X_C}\},\{\overline{P_B-P_C-X_A}\},\{\overline{P_C-P_A-X_B}\}$. But $\{X_A,X_B,X_C\}$ lies on $\{BC,CA,AB\}$ respectively $\implies \{P_AP_B\cap AB=X_C\},\{P_BP_C\cap BC=X_A\},\{P_AP_C\cap AC=X_B\}$. But By Monge D-Albert Circle Theorem we get that $\overline{X_A-X_B-X_C}$. So, $\{\triangle ABC,\triangle P_AP_BP_C\}$ are perspective pairs of triangle. So, By Desargues Theorem we get that $AP_A,BP_B,CP_C$ are concurrent.

The problem is essentially the following lemma. We allow the used of signed radii, so for example the insimilicenter of a circle of positive radius and one of negative radius is simply the exsimilicenter of the two circles with radii given by the absolute values of their signed values. Lemma: Suppose we have circles $(O_i)$ for $i\in\{1,2,3,4\}$, and let $I_{ij}$ be the insimilicenter of $(O_i)$ and $(O_j)$. Then, the lines $I_{12}I_{34}$, $I_{13}I_{24}$, $I_{14}I_{23}$ are concurrent. Proof: Let the (possibly negative) radius of $(O_i)$ be $1/r_i$ for each $i$. Let $w_i=1/r_i$. Note then that \[I_{ij} = \frac{w_i}{w_i+w_j} O_i + \frac{w_j}{w_i+w_j} O_j.\]Note that \[\frac{\sum_{i=1}^4 w_iI_i}{\sum_{i=1}^4 w_i} = \frac{w_1+w_2}{w_1+w_2+w_3+w_4}I_{12}+\frac{w_3+w_4}{w_1+w_2+w_3+w_4}I_{34},\]so \[\frac{\sum_{i=1}^4 w_iI_i}{\sum_{i=1}^4 w_i}\in I_{12}I_{34}.\]The point on the left side is symmetric in $i$, so we have the desired result. $\blacksquare$ The problem now follows by taking the four circles to be $\omega_A$, $\omega_B$, $\omega_C$, and the incircle but with negative radius.

Let $Q_A$ be the exsimilicenter of $\omega_B$ and $\omega_C$ and similarly define $Q_B$ and $Q_C.$ By Monge, $(P_A,P_B,Q_C),$ $(P_B,P_C,Q_A),$ $(P_C,P_A,Q_B),$ and $(Q_A,Q_B,Q_C)$ are collinear. Hence, $\triangle ABC$ and $\triangle P_AP_BP_C$ are perspective and we are done by Desargues. $\square$