Let $X=AD \cap BC$, and suppose $T$ is the foot of perpendicular from $X$ to $EK$. Also let $E'$ be the inverse of $E$ in $\omega$, and note that $EE'PRTX$ is cyclic. Since $K$ is the center of $\odot (PQRS)$, so by Brokard's Theorem, $T$ is the inverse of $E$ in $\odot (PQRS)$. This gives $$KP^2=KT \cdot KE=KQ^2 \Rightarrow (P,R;T,E)=-1$$But then we get that $$(P,R;T,E)=-1=(B,A;E'X \cap AB,E) \stackrel{X}{=} (P,R;E',E) \Rightarrow T \equiv E'$$Thus, $EK \perp E'X$ directly gives that $E,K,O$ are collinear. $\blacksquare$

A synthetic solution: Let $O_1$ and $O_2$ be the circumcenters of the triangles $EBC$ and $EAD$ respectively. From perpendicular bisectors of line segments $BC$ and $AD$, we have $O_1O \perp BC$ and $OO_2 \perp AD$, which also means $EO_1 \parallel OO_2$ and $EO_2 \parallel OO_1$. As a result, $EO_1OO_2$ forms a parallelogram. Let $X$ denote the midpoint of $O_1O_2$. We see that $E,X,O$ are collinear. After some angle chasing, we get that the (degenerate) quadrilaterals $EBSO_1$ and $EDQO_2$ are similar,so we have $\frac{EO_1}{O_1S}=\frac{EO_2}{O_2Q}$i which gives us that $O_1O_2 \parallel QS$ As $X$ and $K$ are the midpoints of these parallel line segments $O_1O_2$ and $QS$, we deduce that $E,X,K$ are collinear from the converse of Thales Theorem,. So, we have $E,X,K,O$ are collinear, which also implies $E,K,O$, as the desired result.

Denote $BC \cap AD = F$, $BD\cap AC = T_1$, $SQ\cap PR = T_2$ Polar line of E written $(ABCD)$ is line $BT_1$ and polar line of E written $(PQRS)$ is $FT_2$ by Brocard. If $BT_1T_2$ are collinear we are done since $EO$ and $EK$ is perpendicular to this line. Denote $X_1$ and $Y_1$ foot of the perpendicular from $T_1$ to $BC$ and $AC$. Similarly define $X_2$ and $Y_2$. We will show $\frac{T_1X_1}{T_1Y_1}=\frac{T_2X_2}{T_2Y_2}$ for the linearity but from similar triangles $\frac{T_1X_1}{T_1Y_1}=\frac{BC}{AD}=\frac{EP}{ER}=\frac{SP}{RQ}=\frac{T_2X_2}{T_2Y_2}$ and we are done