From the radical axis theorem on the circles $(ABC)$, $(BCP)$ and $\omega_A$, the tangent lines to $\omega_A$ passing through $P$ and $A_1$ intersect on the line $BC$, let us say at point $D$. From the radical axis theorem on the circles $(ABC)$, $(BCP)$ and $\Gamma_A$, we see that the line passing through $A_2$ and tangent to the circle $(ABC)$ passes through $D$ as well. Therefore, $D$ is the pole of the line $A_1A_2$. Define the points $E, F$ analogously. Looking at the similarity $DBP \sim DPC$ we have

$$\frac{BP}{PC} = \frac{DB}{DP} = \frac{DP}{DC} \quad \text{and hence} \quad \left(\frac{BP}{PC}\right)^2 = \frac{DB}{DC}$$

$$\left(\frac{CP}{PA}\right)^2 = \frac{EC}{EA} \quad \text{and} \quad \left(\frac{AP}{PB}\right)^2 = \frac{FA}{FB}$$
Therefore, by Menelaus' Theorem we get that the points $D, E, F$ are collinear. Thus, from La Hire Theorem we can deduce that the lines $A_1A_2$, $B_1B_2$ and $C_1C_2$ are concurrent. Let the common point of these lines be $M$. Since $M$ is the pole of the line $\overline{DEF}$ we get $OM \perp DE$. Therefore, in order to complete the solution we should prove that $O, M, P$ are collinear. Hence it suffices to show that $OP$ is perpendicular to $DE$. Let $r$ be the circumradius of the triangle $ABC$. From the power of the point $D$ with respect to the circle $(ABC)$ we have $DP^2 = DB \cdot DC = DO^2 - r^2$. Similarly, from the power of the point $E$ with respect to the circle $(ABC)$ we get $EP^2 = EO^2 - r^2$. Hence $DP^2 + EO^2 = DO^2 + EP^2$ and we get $OP \perp DE$. We are done.