Solution:
Let $Z$ be the point where $PY$ intersects $AX$. The problem asks us to prove that $AZ = ZX$.

Since $\angle BPC = \angle BQC = 90^\circ$ we conclude that $BPQC$ is a cyclic quadrilateral. Hence $BPYQC$ is a cyclic pentagon. A quick angle chase gives us:

$$\angle ZPA = 180^\circ - \angle BPY \\ \qquad = \angle BCY \\ \qquad = 180^\circ - \angle XCY \\ \qquad = \angle XAY \\ \qquad = \angle ZAY.$$

Hence triangles $\triangle ZPA$ and $\triangle ZAY$ are similar ($\angle Z$ is shared). Therefore $\frac{ZP}{ZA} = \frac{ZA}{ZY}$, and hence $ZA^2 = ZP \times ZY$.

Another angle chase gives us:

$$\angle XPZ = \angle QPY \\ \qquad = \angle QCY \\ \qquad = \angle ACY \\ \qquad = \angle AXY \\ \qquad = \angle ZXY.$$

Hence triangles $\triangle ZPX$ and $\triangle ZXY$ are similar ($\angle Z$ is shared). Therefore $\frac{ZP}{ZX} = \frac{ZX}{ZY}$ and hence $ZX^2 = ZP \times ZY$.

Putting this together gives us:

$$ZA^2 = ZP \times ZY = ZX^2.$$

Hence $ZA = ZX$, and therefore $Z$ is the midpoint of $AX$.

Alternative Solution (outline):

First we will embed the diagram in the Argand plane, such that point $B$ is represented by the complex number $b = -1$ and point $C$ is represented by the complex number $c = 1$. Lower case letters will always denote the complex number representing the corresponding upper case letter (so $a$ is the complex number representing point $A$ and $x$ is the complex number representing point $X$, etc). We will endeavour to find expressions for all the points in the diagram in terms of $p$ and $q$.

Since $\angle BPC = \angle BQC = 90^\circ$, we know that points $P$ and $Q$ both lie on the unit circle. So $p\bar{p} = q\bar{q} = 1$. Therefore the circumcircle of triangle $PQC$ is the unit circle and thus $y\bar{y} = 1$ too. Since point $A$ is the intersection of chords $BP$ and $CQ$, we can compute $a$ using the formula for the intersection of two chords.

$$a = \frac{cq(b + p) - bp(c + q)}{cq - bp} = \frac{q(-1 + p) - (-1)p(1 + q)}{q - (-1)p} = \frac{2pq + p - q}{p + q}$$

$$\Rightarrow \qquad \bar{a} = \frac{2\bar{p}\bar{q} + \bar{p} - \bar{q}}{\bar{p} + \bar{q}} = \frac{(2\bar{p}\bar{q} + \bar{p} - \bar{q})pq}{(\bar{p} + \bar{q})pq} = \frac{2 + q - p}{q + p}$$

$$\mathrm{note:}\qquad \frac{1 - a}{1 - \bar{a}} = \frac{1 - \frac{2pq + p - q}{p + q}}{1 - \frac{2 + q - p}{q + p}} = \frac{(p + q) - (2pq + p - q)}{(p + q) - (2 + q - p)} = \frac{2q - 2pq}{2p - 2} = -q.$$

Similarly $X$ is the intersection of chords $PQ$ and $BC$, so

$$x = \frac{pq(b + c) - bc(p + q)}{pq - bc} = \frac{pq(0) - (-1)(p + q)}{pq - (-1)} = \frac{p + q}{pq + 1}$$

Now we have formulas for $a$ and $x$ in terms of $p$ and $q$. Next we will use the fact that $AYCX$ is cyclic to find a formula for $y$ in terms of $p$ and $q$. $AYCX$ being cyclic is equivalent to $\angle CAY = \angle CXY$. This is equivalent to

$$\left(\frac{c - a}{\bar{c} - \bar{a}}\right) / \left(\frac{y - a}{\bar{y} - \bar{a}}\right) = \left(\frac{c - x}{\bar{c} - \bar{x}}\right) / \left(\frac{y - x}{\bar{y} - \bar{x}}\right)$$

To simplify this, first recall that $\frac{c - a}{\bar{c} - \bar{a}} = \frac{1 - a}{1 - \bar{a}} = -q$. Also, since $c = \bar{c}$ and $x = \bar{x}$ ($c$ and $x$ are real numbers) the $\frac{c - x}{\bar{c} - \bar{x}}$ factor is 1. Furthermore, since $y\bar{y} = 1$ we can replace $\bar{y}$ with $y^{-1}$. Thus the equation for $AYCX$ being cyclic becomes:

$$(-q) / \left(\frac{y - a}{y^{-1} - \bar{a}}\right) = 1 / \left(\frac{y - x}{y^{-1} - x}\right).$$

From here, we can multiply out the denominators, expand the brackets and collect like terms to get a quadratic in $y$.

$$(q\bar{a} + x)y^2 - (q\bar{a}x + q + xa + 1)y + (qx + a) = 0.$$

Now (using $\frac{1 - a}{1 - \bar{a}} = -q$) we can get $(q\bar{a} + x + qx + a) = (q\bar{a}x + q + xa + 1)$. Thus the quadratic factorises as:

$$(y - 1)\Big((q\bar{a} + x)y - (qx + a)\Big) = 0.$$

Since $Y$ and $C$ are distinct points, we know $y \neq 1$ and so we finally get a formula for $y$

$$(q\bar{a} + x)y - (qx + a) = 0\qquad \Longrightarrow \qquad y = \frac{qx + a}{q\bar{a} + x}$$

We can now substitute our formulas for $a$ and $x$ ($a = \frac{2pq + p - q}{p + q}$ and $x = \frac{p + q}{pq + 1}$) into this expression to find $y$ in terms of $p$ and $q$. After some algebraic simplification this yields:

$$y = \frac{2p^2q + (q + 1)p + q^2 - q}{(1 - q)p^2 + (q + q^2)p + 2q}.$$

Let $M$ be the midpoint of $AX$. So

$$m = \frac{a + x}{2} = \frac{\frac{2pq + p - q}{p + q} + \frac{p + q}{pq + 1}}{2} = \frac{(pq + 1)(2pq + p - q) + (p + q)^2}{2(p + q)(pq + 1)}.$$

$$\overline{m} = \frac{(pq + 1)(2 + q - p) + (p + q)^2}{2(p + q)(pq + 1)}.$$

$$p\overline{m} - 1 = \frac{(p - 1)((1 - q)p^2 + (q + q^2)p + 2q)}{2(p + q)(pq + 1)}$$

$$y(p\overline{m} - 1) = \frac{(p - 1)(2p^2q + (q + 1)p + q^2 - q)}{2(p + q)(pq + 1)}$$

$$y(p\overline{m} - 1) + m = \frac{(p - 1)(2p^2q + (q + 1)p + q^2 - q)}{2(p + q)(pq + 1)} + \frac{(pq + 1)(2pq + p - q) + (p + q)^2}{2(p + q)(pq + 1)}$$

$$= \frac{(p - 1)(2p^2q + (q + 1)p + q^2 - q) + (pq + 1)(2pq + p - q) + (p + q)^2}{2(p + q)(pq + 1)}$$

$$= \frac{2p^3q + 2p^2q^2 + 2p^2 + 2pq}{2(p + q)(pq + 1)}$$

$$= p.$$

We have shown that $y(p\overline{m} - 1) + m = p$. Hence

$$m = p + y - py\overline{m}.$$

Which interpreted geometrically means that point $M$ lies on chord $PY$ of the unit circle.