Wonder why no solution has been posted yet, but whatever. Call the midpoint of minor arc $AB$ and $AC$ of the circumcircle $D$ and $E$ respectively. Also, let $Y$ be the intersection of $AP$ and $I_1I_2$ and $X$ be the intersection of $ZP$ and the circumcircle. (In particular, $Y$ is the insimilicenter of the two incircles.) Since $BC$ is a external common tangent to the incircles, $Z$ must be their exsimilicenter and thus $(Z,Y;I_1,I_2)=-1$. Projecting this with pencil $P$ onto the circumcircle, we have $(X,A;D,E)=-1$. Since $A,D,E$ are all fixed, $X$ must also be independent to $P$ and we are done.

Note that $X$ is mixtilinear point.