Excuse me because I am not Darij, but I'll try to prove this
Let $E$ be second intersection point of line $BP$ and circle. Then
$PB\cdot PE=PA^2=PB\cdot PD$ so $PE=PD$.$F$ is intersection point of
line $BD$ and circle. Now, $\triangle PEF \cong \triangle PBD$, so
quadrilateral $BFED$ is a trapezoid.
$AC$ is a polar of point $P$ wrt circle so lines $BD,FE$ and $AC$
are concurrent. Line passing through P and midpoints of $BF$ and
$ED$ (remember $PE=PD$ and $PB=PF$) is also passing through center
of circle (because $BFDE$ is a trapezoid), so it's passing through
midpoint of $AC$ and through interection point of diagonals $BD$ and
$EF$. So intersection point of lines $AC,BD$ and $EF$ is midpoint of
$AC$. The end
If you don't understand just ask
bye