$\sqrt{bc}$ inversion kills. Make a $\sqrt{bc}$ inversion in $\triangle ABC$, so now $T$ goes to $Q$, $I$ goes to $I_A$ (the A-excenter) so $AQ \cap (ATI_A)=K'$ but since $AQ, AT$ are isogonal w.r.t. $\angle BAC$ and $AI$ bisects that angle we have that in $(ATI_AK')$ the arcs $I_AK'$ and $I_AT$ are equal so their lenghts are equal meaning that $I_AK'$ equals to the exradius hence $K'$ lies in the A-excircle so inverting back we have that $K$ lies in the A-mixtilinear incircle, but note that $Q$ is the tangent point between the A-mixtilinear incircle and $\Omega$ so $(DEQ)$ is the A-mixtilinear incircle and that means $DKEQ$ cyclic thus we are done

Let M be the intersection point of incircle and BC. ∵BC∥PQ, BM=CT, ∴QMTP is an isosceles trapezoid. ∴∠QMC=π-∠QPT=π-∠QCA=∠APQ. ∵∠BAQ=∠MCQ, ∴∆QBA~∆QMC. So Q is the intersection point of mixtilinear circle and (ABC). According to the Mannheim theorem, (QDE) is the mixtilinear circle. Using angle chasing, we know that the north point of(ABC) is on the line QI. As (QDE) and (ABC) are homothetic about point Q, QI∩(QDE)=the north point of (QDE). Let this point be K'. Find the line through K' tangent to (QDE), and this line intersects AB,AC at X,Y. XY∥BC. So (QDE) is the excircle of ∆AXY, AK' is the A-Nagel line of it. And AT is the A-Nagel line of ∆ABC, so K=K‘. And we are done.

Let $M$ be the midpoint of arc $BAC$ and redefine $K$ as the second intersection between $QI$ and $(QDE)$. Since $AP$ and $AQ$ are isogonal in $\angle BAC$, it follows that $(QDE)$ is the $A$-Mixtilinear Incircle. Thus, a well-known lemma implies $M \in QI$. Because $Q, K, M$ are collinear, the positive homothety at $Q$ taking $(QDE)$ to $(ABC)$ maps $K$ to $M$. Now, since the tangent to $(ABC)$ at $M$ is parallel to $BC$, we know the positive homothety mapping $(ABC)$ to the $A$-excircle takes $M$ to $T$. Hence, the positive homothety taking $(QDE)$ to the $A$-excircle maps $K$ to $T$. But the center of this homothety is clearly $A$, so we have $K \in AT$, which finishes. $\blacksquare$

$\angle QAB=\angle TAC$ so $(QDE)$ is mixtilinear circle and is tangent to $AC,AB,(ABC)$ let $K_1$ be intersection of $(QDE),IQ$ : $\frac{sin \angle K_1AE}{sin \angle K_1AD} = \frac{K_1E}{DK_1} . \frac{sin \angle K_1DA}{sin \angle K_1EA}$ and because mixtilinear is tangent to $AC,AB,(ABC)$ and $QIEC,DBQI$ are cyclic : $\frac{sin \angle K_1AE}{sin \angle K_1AD} = \frac{K_1E^2}{DK_1^2} = \frac{sin^2 C/2}{sin^2 B/2}$ Just we should show that : $\frac{sin \angle KAE}{sin \angle KAD} = \frac{sin^2 C/2}{sin^2 B/2}$ Lemma 1 : if $ABCD$ be a cyclic quardital and $X$ be intersection $AC,BD$ then : $\frac{AX}{CX} = \frac{AB}{BC} . \frac{AD}{BD}$ Lemma 2 : $sin^2 A/2= \frac{(p-b)(p-c)}{bc}$ now using Lemma 1, Lemma 2 : $\frac{sin \angle KAE}{sin \angle KAD} =\frac{PC}{BP}= \frac{p-b}{p-c} . \frac{c}{b} = \frac{sin^2 C/2}{sin^2 B/2}$ $CT=p-b,BT=p-c$ So $K_1=K \Rightarrow DKEQ$ is cyclic $\blacksquare$

Let $Q'$ be intersection of $BDI$ with $ABC$. Clearly $CEIQ'$ is cyclic. we will prove $Q'$ is $Q$ $\angle DQ'E = \angle DQ'I + \angle EQ'I = \angle DBI + \angle ECI = \angle 90 - \frac{A}{2} = \angle ADE = \angle AED$ so $DQ'E$ is tangent to $AD$ and $AE$. Note that $Q'A$ is symmedian in $DQ'E$ so $\angle AQ'D = \angle EQ'I = \angle ECI$. Let $l$ be tangent to $\Omega$ and $R$ an orbitrary point on $l$. Now we have $\angle DQ'R = \angle AQ'R - \angle AQ'D = \angle ACQ' - \angle ACI = \angle ICQ' = \angle IEQ' = \angle DEQ'$ so $l$ is tangent to $DEQ'$ so $Q'ED$ is $A$-Mixtilinear. $\frac{BQ'}{CQ'} = \frac{BD}{CE} = \frac{\frac{BD}{DI}}{\frac{CE}{EI}} = \frac{\frac{\sin{BQ'D}}{\sin{DQ'I}}}{\frac{\sin{CQ'E}}{\sin{EQ'I}}} = \frac{\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}}^2 = \frac{p-b}{p-c} . \frac{c}{b} = \frac{\sin{CAT}}{\sin{BAT}} = \frac{CP}{BP}$ so $Q'$ is $Q$. Now Assume $QDE$ meets $IQ$ at $K'$. $\frac{\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}} = \frac{K'E}{K'D} = \frac{\sin{K'AE}}{\sin{K'AD}} . \frac{\sin{K'DA}}{\sin{K'EA}} \implies \frac{\sin{K'AE}}{\sin{K'AD}} = \frac{\sin{\frac{C}{2}}}{\sin{\frac{B}{2}}}^2 = \frac{p-b}{p-c} . \frac{c}{b} = \frac{\sin{CAT}}{\sin{BAT}} = \frac{\sin{KAE}}{\sin{KAD}} \implies \angle KAE = \angle K'AE , \angle KAD = \angle K'AD$ so $K'$ and $K$ are the same so $DKEQ$ is cyclic. (I think removing "Mixtilinear" from the title would be a good idea btw)