Solution to (a) Note that $\angle AQP = \angle ABP = \angle ADE$ and $\angle APQ = \angle ACQ = \angle AED$, so we have a spiral similarity $\triangle ADE \sim \triangle AQP$. Therefore, lines $PE$ and $QD$ meet at the second intersection of $\omega$ and $\Omega$ other than $A$.

Solution to (b) using angle chasing Let $L$ be the reflection of $H$ across $\overline{AB}$, which lies on $\Omega$.
Claim — Points $L, D, P$ are collinear.
Proof. This is just angle chasing:
$$\begin{aligned} \angle CLD &= \angle DHL = \angle DHA + \angle AHL = \angle DEA + \angle AHC \\ &= \angle ADE + \angle CBA = \angle ABP + \angle CBA = \angle CBP = \angle CLP. \end{aligned} \quad \square$$

Now let $K \in \omega$ such that $DHKE$ is an isosceles trapezoid, i.e. $\angle BAH = \angle KAE$.
Claim — Points $D, K, P$ are collinear.
Proof. Using the previous claim,
$$\angle KDE = \angle KAE = \angle BAH = \angle LAB = \angle LPB = \angle DPB = \angle PDE. \quad \square$$
By symmetry, $\overline{QE}$ will then pass through the same $K$, as needed.

Solution to (b) by orthogonal circles (found by contestants) We define $K$ as in the previous solution, but do not claim that $K$ is the desired intersection. Instead, we note that:
Claim — Point $K$ is the orthocenter of isosceles triangle $APQ$.
Proof. Notice that $AH = AK$ and $BC = PQ$. Moreover from $\overline{AH} \perp \overline{BC}$ we deduce $\overline{AK} \perp \overline{PQ}$ by reflection across the angle bisector.
In light of the formula "$AH^2 = 4R^2 - a^2$", this implies the conclusion. $\square$
Let $M$ be the midpoint of $\overline{PQ}$. Since $\triangle APQ$ is isosceles, $\overline{AKM} \perp \overline{PQ}$ and we conclude that $MK \cdot MA = MP^2$. So the circle with diameter $\overline{PQ}$ is orthogonal to $\omega$. Combined with (a), this implies the result by Brokard theorem.

Solution to (b) by complex numbers Let $M$ be the arc midpoint of $\widehat{BC}$. We use the standard arc midpoint configuration. We have that
$$A = a^2, \ B = b^2, \ C = c^2, \ M = -bc, \ H = a^2 + b^2 + c^2, \ P = \frac{a^2c}{b}, \ Q = \frac{a^2b}{c},$$
where $M$ is the arc midpoint of $\widehat{BC}$. By direct angle chasing we can verify that $\overline{MB} \parallel \overline{DH}$. Also, $D \in \overline{AB}$. Therefore, we can compute $D$ as follows.
$$d + a^2b^2\bar{d} = a^2 + b^2 \quad \text{and} \quad \frac{d-h}{\bar{d}-\bar{h}} = -mb^2 = b^3c \implies d = \frac{a^2(a^2c+b^2c+c^3-b^3)}{c(bc+a^2)}.$$
By symmetry, we have that
$$e = \frac{a^2(a^2b + bc^2 + b^3 - c^3)}{b(bc + a^2)}.$$
To finish, we want to show that the angle between $\overline{DP}$ and $\overline{EQ}$ is angle $A$. To show this, we compute $\frac{d-p}{e-q} \big/ \frac{d-p}{e-q}$. First, we compute
$$\begin{aligned} d - p &= \frac{a^2(a^2c + b^2c + c^3 - b^3)}{c(bc + a^2)} - \frac{a^2c}{b} \\ &= a^2 \left( \frac{a^2c + b^2c + c^3 - b^3}{c(bc + a^2)} - \frac{c}{b} \right) = \frac{a^2(a^2c - b^3)(b-c)}{bc(bc + a^2)}. \end{aligned}$$
By symmetry,
$$\frac{d-p}{e-q} = \frac{a^2c - b^3}{a^2b - c^3} \implies \frac{d-p}{e-q} \big/ \frac{d-p}{e-q} = \frac{a^2b^3c}{a^2bc^3} = \frac{b^2}{c^2}$$
as desired.