From well known spiral sim config, we see that $P$ (which we rename to $S$ in what follows) is the centre of spiral similarity mapping $ED \mapsto BC$. Moreover, we know that $SO \equiv SU$ bisects $\angle BSD$ and $\angle ESC$. Now let $SO$ meet circle $(O)$ again at a point $I$, such that $I$ lies strictly inside segment $PO$. Now $OB = OD = OI$ along with $SI$ bisecting $\angle BSD$. Since $O$ is the arc midpoint, it follows that $I$ is the incentre of $\triangle SBD$. Similarily, $I$ is the incentre of $\triangle SEC$ and we're done. $\square$