Lemma : $ABC$ is a triangle with incenter $I$ ; $A'$ the midpoint of arc $BC$ not containing $A$.Let the bissectors of $\angle AA'B,\angle AA'C$ cut respectively $BA,CA$ at $U,V$ then $I,U$ and $V$ are collinear and $UV\parallel BC$ Proof : Let the parallel to $BC$ through $I$ cut $AB,AC$ at $ U',V'$ ; it ' s clear that $IU'B,IV'C$ are isoceles besides $A'I=A'B=A'C$ we get $A'IU=A'BU$ thus $A'U' $ is bisector of $ \angle IA'B$ hence $U'=U$ idem $V'=V$ Back to the problem : Let $B'$ the midpoint of arc $BC$ not containing $B$; $Q=BB\cap A_0C_0$ we need to show that $Q=P$ Lemma ascertains that if $B'A_0\cap BC=V,B'C_0\cap BA=U$ then $I\in UV\parallel AC$. applying pascal to $BBCC_0A_0B'$we get $Q-V-I$ are collinear which closes the proof .

No need to use harsh theory. Angle chasing gives $C_0I = C_0B$ and $A_0I = A_0B$ - so $PC_0A_0$ is the perpendicular bisector of $BI$, whence $\triangle PC_0B \cong \triangle PC_0I$. Hence $\angle PBC_0 = \angle PIC_0 = \angle ACI$ (the latter is from the line parallel to $AC$) and $\angle ABP = \angle PBC_0 + \angle ABC_0 = 2\angle ACI = \angle ACB$, which solves the problem.

AC₀BA₀C is cyclic.Firstly we note that <ACC₀=<BCC₀=α and <BAA₀=<CAA₀=β.From cyclic <C₀BA=<C₀AB=<C₀A₀A=<C₀A₀B=α similarly <A₀BC=<A₀C₀C=<BC₀A₀=β.PI parallel to AC and we get from this C₀IP=α.We get <A₀PI=β-α and <CBI=90-α-β from angle chasing.Then we get A₀C₀ perpendicular to BI and bisects BI.Then we get PB=PI.So we get <BPA₀=β-α and <PBC₀=α. From <PBC₀=<BA₀C₀ we get PB touch k.We are done.

Attachments: