Oops idk what I was thinking when I wrote this....

We'll use two well-known mixtilinear incircle lemmas: If it tangent to $AB,AC,(ABC)$ at $D,E,Q$, then $I\in DE$ and $QI$ goes through midpoint of the arc $BAC$. Note that $FDE$ is homothetic at $A$ with tangential triangle of $A$-excircle, so $(FDE)$ is inscibed in $\angle A$. Fact that $I\in DE$ makes it mixtilinear incircle, so it tangent to $(ABC)$, say at $Q$. By above homothety, tangent at $F$ is parallel to $BC$, so $F\in QI$. Hence $P\equiv Q$ and result follows.

Let $P'$ be the $A$-mixtilinear touch point with the circumcircle of $\triangle ABC$. Let $P'I\cap (P'DE) = F'$. Now note that $\angle ABI = \frac{\angle B}{2} = \angle DP'I = \angle ADF' \implies DF'\parallel BI$. Similarly, $\angle ACI = \frac{\angle C}{2} = \angle F'P'E = \angle F'EA \implies EF'\parallel CI$. So that means $F = F'$. So $P = P'$. So $\omega$ is the $A$-mixtilinear incircle of $\triangle ABC$ and now it is evident that $O,K,P$ are collinear as desired. $\square$ @below Sorry, I meant to say $P'$ instead of $T$. Edited.

hakN wrote: Let $P'$ be the $A$-mixtilinear touch point with the circumcircle of $\triangle ABC$. Let $P'I\cap (P'DE) = F'$. Now note that $\angle ABI = \frac{\angle B}{2} = \angle DTI = \angle ADF' \implies DF'\parallel BI$. Similarly, $\angle ACI = \frac{\angle C}{2} = \angle F'TE = \angle F'EA \implies EF'\parallel CI$. So that means $F = F'$. So $P = P'$. So $\omega$ is the $A$-mixtilinear incircle of $\triangle ABC$ and now it is evident that $O,K,P$ are collinear as desired. $\square$ Excuese me，what is point T? An arbitrary point on $\omega$?

(DFE) is the A-mixtilinear incircle. So the result follows directly by using its properties.

$(DEF)$ is just the A-mixtilinear incircle and this just finishes it. To show $F$ lies on it ,we use cyclicity of $(DIPB)$ or $(IECP)$. Quite direct.

Hmmm hopefully I'm not high this time. Note that $\triangle ADE$ is isosceles, $\angle ADF = \angle ABI = \dfrac 12 \angle B \implies \angle FDE = \dfrac 12 \angle C$ and similarly $\angle AEF = \dfrac 12 \angle C$ and $\angle FED = \dfrac 12 \angle B$ so $\omega$ is tangent to $AD,AE$. $$\angle DPI = \angle DEF = \dfrac 12 \angle B = \angle DBI$$So $DBPI$ is cyclic, similarly $ECPI$ is cyclic. $$\angle BPC = \angle BPI + \angle IPC = \angle IDA + \angle IEA = 180^\circ - \angle A$$So $P \in (ABC)$ and combining the tangency fact from above we conclude that $P$ is the A-Mixtilinear touchpoint with $(ABC)$ which finishes the problem.