Let $I$ be the incenter and $I_a$ the $A$-excenter of triangle $ABC$ and $J$ the $C$-excenter of triangle $CDE$. $J$ lies on $\omega$, since
$$\angle EJD = 90^\circ - \frac{1}{2}\angle ECD = \angle CDE.$$
We claim that $A$, $J$, and $Z$ are collinear. Point $T$ lies on $AC$ such that $IT \parallel DE$. To prove our claim, we just need to show that $\frac{IJ}{IT} = \frac{I_a Z}{I_a C}$, since $IT \parallel CI_a$, $IJ \parallel I_a Z$ and $II_a$ passes through $A$. Notice that
$$\angle ITE = \angle DEC = \angle EDC = \angle I_a CY \implies \triangle IET \sim \triangle I_a YC \implies \frac{IE}{IT} = \frac{I_a Y}{I_a C},$$
hence $\frac{IJ}{IT} = \frac{I_a Z}{I_a C}$, since $IE = IJ$ and $I_a Y = I_a Z$. So the claim is proved. Let $BC$ touches $A$-excircle at $Y$. The extensions of sides of quadrilateral $AEDB$ are tangent to $A$-excircle, therefore
$$AE + ED = AB + BD \implies ED = 2BD. \quad (1)$$
Yielding $DX = DE$, and it is obvious that $DY = DZ$ so quadrilateral $YZXE$ is an isosceles trapezoid and $Y$ lies on the circumcircle of triangle $XZE$. Hence
$$\angle EKY = \angle EZY = \frac{1}{2}\angle EDY = \angle EDM,$$
where $M$ is the midpoint of arc $ED$ (not containing $J$), so $KY$ passes through $M$. Let $N$ be the midpoint of segment $ED$. Notice that $\triangle CDI \sim \triangle CND$, hence
$$\frac{DN}{CD} = \frac{ID}{IC} \stackrel{(1)}{\implies} \frac{BD}{CD} = \frac{IJ}{IC} \implies JB \parallel ID.$$

therefore $\angle YJD = \angle BKD$, since $\angle BJY = \angle BKY$. According to our assumption, $J$ lies on the angle bisector of $\angle YDZ$ and $YD = DZ$. It yields that $JD$ is the perpendicular bisector of $YZ$ and $\angle YJD = \angle ZJD$. Hence $\angle ZJD = \angle BKD$ and the result follows.