What i got till now. Let $D,E,F$ be the feet of angle bisectors of $\angle APB,\angle CPB,\angle CPA$ The condition implies $AD=AF$,$BD=BD$,$CE=CF$ thus $DEF$ is the intouch triangle of $ABC$ $\implies$ $AP:BP:CP=AD:BE:CF$. Let $D^{'},E{'},F^{'}$ be the harmonic conjugates of $D,E,F$ wrt $AB,BC,AC$ then $P$ is the intersection of the Apollonius circles of with radius $DD^{'}$,$EE^{'}$,$FF^{'}$(which gives a way to construct the point). We only need that they share the center of the incircle becaus $r=\frac{R^{2}-OI^{2}}{2R}$

Let $O$ be the circumcenter, $I$ the incenter, $DEF$ the intouch triangle, $D',E',F'$ harmonic conjugates of $D,E,F$ wrt $BC,CA,AB$. Then since $AP:BP:CP=AE:BF:CD$, $P$ is the intersection of the Apollonius circles with diameter $DD',EE',FF'$. If $X,Y,Z$ are midpoints of $DD',EE',FF'$, then $XYZ$ is the radical axis of the incircle and circumcircle and point $P$, so $I,O,P$ are collinear (and perpendicular to $XYZ$). Let $M,N$ be the feet of perpendiculars from $P,O$ onto $AB$, and $M',N',L$ be the feet of perpendiculars from $P,O,I$ onto $A'B'$. Then since $\triangle ABP\sim\triangle B'A'P$ and $P,O,I$ are collinear, we have $ANDMBP\sim B'N'LM'A'P$, thus $PL$ bisects $\angle B'PA'$, so $D,P,L$ are collinear. Since $ID\perp AB,IL\perp A'B'$ and $\angle BDP=\angle A'LP$, we have $\angle IDP=\angle ILP\implies ID=IL$. Thus $A'B'$ is tangent to the incircle at $L$ and the result follows.