Czech-Polish-Slovak Match

Assume that the lines $AI$, $BI$, $CI$, $DI$ meet the circumcircle of the quadrilateral $ABCD$ at $E$, $F$, $G$, $H$, respectively. Since the lines $AI$, $BI$, $CI$, $DI$ are the bisectors of the respective angles of the quadrilateral $ABCD$, the lines $EG$ and $FH$ are the diameters of the circumcircle of $ABCD$. Thus $EG$ and $FH$ meet at $O$. Denote by $X$ the point of intersection of $EB$ and $CH$. Using Pascal's theorem for the hexagon $ACHDBE$ we see that $P$, $X$ and $I$ are collinear. Once again Pascal's theorem applied to the hexagon $GCHFBE$ yields that $O$, $X$ and $I$ are collinear. Thus $O$, $I$ and $P$ are collinear, as desired.