Solution:

Construct the tangent line to $\Gamma$ at $P$. Note that this line is also tangent to the circle through points $C$ and $P$ with centre $O$. Also construct point $E$ on this tangent line to the right of $P$. Note that $\angle EPO = 90^\circ$ and $\angle OCA = 90^\circ$ because the radii and tangents are perpendicular.

Let $x = \angle PBQ$

$\angle EPQ = x$ (by alternate segment theorem)

$\angle OPQ = x - 90^\circ$ (because $\angle EPO = 90^\circ$)

$\angle QCO = 180^\circ - \angle OPQ$ (because opposite angles in a cyclic quad $= 270^\circ - x$ are supplementary)

$\angle ACQ = 360^\circ - \angle QCO - \angle OCA$ (angles around point $C$ are $360^\circ$)

$= 360^\circ - (270^\circ - x) - 90^\circ$

$= x$.

We also have $\angle QPB = \angle QAB$ (angles subtended by chord $QB$) in cyclic quad $QAPB$.

Therefore we have similar triangles

$$\triangle QBP \sim \triangle QCA \qquad (\angle PBQ = \angle ACQ \mathrm{~and~} \angle QPB = \angle QAC)$$

Hence $\angle AQC = \angle PQB$. Therefore

$$\angle AQP = \angle AQC + \angle CQP = \angle PQB + \angle CQP = \angle CQB.$$