Let the lines $\ell_{B}$ and $\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.

Claim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic.

Proof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have
$$\text{(O A P, O C P )= (O C O, O A O )= (C P, A P)= (C B, A B)= (O A L B , O C L B ) .}$$
Denote the circle through $L_{B}, P, O_{A}$ and $O_{C}$ by $\omega_{B}$. Define the circles $\omega_{A}$ and $\omega_{C}$ similarly.

Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$.

Proof. From cyclic quadruples of points in the circles $\omega_{B}$ and $\omega_{C}$, we have
$$\text{(L C L A , L C P ) = (L C O B , L C P )= (O A O B , O A P ) = (O A O C , O A P )= (L B O C , L B P )= (L B L A , L B P ) .}$$

Claim 3. The points $P, L_{C}$, and $C$ are collinear.

Proof. We have $\text{(P L C , L C L A )= (P L C , L C O B )= (P O A , O A O B )}$. Further, since $O_{A}$ is the centre of the circle $A O P$, $\text{(P O A , O A O B )= (P A, A O)}$. As $O$ is the circumcentre of the triangle $\text{P C A, (P A, A O)= / 2- (C A, C P)= (C P, L C L A )}$. We obtain $\text{(P L C , L C L A )= (C P, L C L A )}$, which shows that $P \in C L_{C}$.

Similarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that
$$\text{(O P, P L A )= (P A, A O)= (P L C , L C L A ),}$$
which means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$.