I'm not particularly good at geometry but I think I've devised one approach to the problem. Without loss of generality let the unit circle of triangle [imath]ABC[/imath] be the circumcircle. Let [imath]A=-i[/imath], [imath]B=a+bi[/imath] and [imath]C=c+di[/imath].
First, let's consider the equation of line through points. Suppose [imath]Z[/imath] is an arbitrary point on the line through [imath]W_1[/imath] and [imath]W_2[/imath]. Since the vector from [imath]W_1[/imath] to[imath]Z[/imath] is a multiple of the vector from [imath]W_1[/imath] to [imath]W_2[/imath], in terms of complex numbers we have [imath]Z-W_1=\lambda(W_2-W_1)[/imath] for some real [imath]\lambda[/imath]. Now [imath]\lambda=\bar{\lambda}[/imath] hence
[math]\frac{Z-W_1}{W_2-W_1}=\frac{\bar{Z}-\bar{W_1}}{\bar{W_2-W_1}}[/math]and any [imath]Z[/imath] satisfying the above equation corresponds to a point on the line through [imath]W_1[/imath] and [imath]W_2[/imath]. Next, consider the line passing through a point [imath]C[/imath] and perpendicular to the line through [imath]W_1[/imath] and[imath]W_2[/imath]. Let [imath]Z[/imath] be on this line. Then the vector from [imath]C[/imath] to [imath]Z[/imath] is perpendicular to the vector from [imath]W_1[/imath] to [imath]W_2[/imath]. In terms of complex numbers, we get [imath]Z-C=i\lambda(W_2-W_1)[/imath] for some real [imath]\lambda[/imath].
Hence [math]\frac{Z-C}{i(W_2-W_1)}=\frac{\bar{Z}-\bar{C}}{-i(\bar{W_2}-\bar{W_1})}[/math].
In the case when [imath]W_1[/imath] and [imath]W_2[/imath] are on the unit circle we have [imath]W_1\bar{W_1}=W_2\bar{W_2}=1[/imath]. Multiplying the numerators and denominators of the right side of the two equations above by [imath]W_1W_2[/imath] we can simplify them as
[math]Z+W_1W_2\bar{Z}=W_1+W_2[/math] and [math]Z-W_1W_2\bar{Z}=C-W_1W_2\bar{C}[/math]
Now since [imath]A[/imath], [imath]B[/imath]and [imath]C[/imath] are points on the unit circle these formulas can be applied immediately to compute [imath]Q[/imath] and [imath]P[/imath].
The line through [imath]C[/imath] perpendicular to [imath]AB[/imath] is given by
[math]Z-AB\bar{Z}=C-AB\bar{C}[/math]
And the line through [imath]0[/imath] and [imath]-i[/imath] is [imath]Z=-\bar{Z}[/imath]. Since these two lines intersect at [imath]Q[/imath], solving this system of equations for [imath]Z[/imath] gives us [imath]Q[/imath]. Hence
[math]Q=\frac{C-AB\bar{C}}{1+AB}[/math]
Similarly, the line through [imath]B[/imath] perpendicular to [imath]AC[/imath] is
[math]Z-AC\bar{Z}=B-AC\bar{B}[/math]
And this line intersects [imath]Z=-\bar{Z}[/imath] at [imath]P[/imath]. Hence
[math]P=\frac{B-AC\bar{B}}{1+AC}[/math]
Now with complex numbers, the orthocenter [imath]H=A+B+C[/imath]. Hence each vertices of the triangle [imath]HPQ[/imath] can be computed and a similar method could now be used to determine [imath]O'[/imath]. Now [imath]M =\frac{B+C}{2}[/imath]and we must show that [imath]M, O'[/imath] and [imath]A[/imath] are colinear. In other words that
[math]\frac{A-O'}{O'-M}=\lambda\in \mathbb{R}[/math]
