Let $A'$, $B'$, $C'$ and $D'$ be the points diametrically opposite $A$, $B$, $C$ and $D$, respectively, on $(O)$, and let $AD' \cap BC' = X$ and $A'D \cap B'C = Y$. Applying Pascal's theorem to cyclic hexagons $ADC'D'CB$ and $DAB'A'BC$ yields that $O$, $P$, $X$ and $Y$ are collinear. Hence it suffices to show that $Q$ lies on $XY$. Let the excircle of $\triangle{EBC}$ be tangent to $BC$ at $R$. Since triangles $\triangle{EBC}$ and $\triangle{EDA}$ are similar, it follows that $AT_2/T_2 D = CR/RB$. Applying the lemma that $BT_1 = CR$, it follows that $AT_2/T_2 D = CR/RB = BT_1 / T_1 C$. Now note that $IT_1$, $BX$ and $CY$ are all perpendicular to $BC$ and hence are parallel. Similarly $JT_2$, $AX$ and $DY$ are all perpendicular to $AD$ and hence are parallel. Now let $IT_1$ intersect $XY$ at $Q'$. It follows that $XQ'/Q'Y =BT_1 / T_1 C = AT_2/T_2 D$, which implies that $Q'T_2$ is parallel to $AX$ and $DY$. Hence $Q'$ lies on $JT_2$ and $Q=Q'$. Hence $X$, $Y$ and $Q$ are collinear, which implies that $O$, $P$ and $Q$ are collinear.

Let $F = BC\cap AD$. Let $M$ be the Miquel Point of $ABCD$. We know that $O,P,M$ are collinear and in particular, $\angle OMF = 90$. So it suffices to show that $FMT_1T_2Q$ is cyclic. It is clear that $FT_1T_2Q$ is cyclic. Now, we will prove that $FMT_1T_2$ is cyclic. Because $\triangle EBC \sim \triangle EDA$, we have $\frac{BT_1}{T_1C} = \frac{AT_2}{T_2D}$. So, spiral similarity centered at $M$ which takes $BC$ to $AD$ also takes $T_1$ to $T_2$. So , $\angle MT_1F = \angle MT_1B = \angle MT_2A = \angle MT_2F \implies MT_1T_2F$ is cyclic. We are done.