Just angle chase.

Let $AD$ intersect smaller circle at $X$ because they are internally tangent there exist a homothety with center $A$ which send points $B \to C$ and $X \to D$ so we get $BX$ $\mid \mid$ $CD$ and we get $\angle AQB = \angle AXB = \angle ADC$ so $ADPQ$ is concyclic.

Construct a tangent to both circles at A. Now by the tan-chord theorem we have ADC and AQB are equal. So ADPQ is cyclic. I haven't checked all the cases but I think it's the same

Here's an Inversive Solution.... Indonesia MO 2018 Day 2 P2 wrote: Let $\Gamma_1, \Gamma_2$ be circles that touch at a point $A$, and $\Gamma_2$ is inside $\Gamma_1$. Let $B$ be on $\Gamma_2$, and let $AB$ intersect $\Gamma_1$ on $C$. Let $D$ be on $\Gamma_1$ and $P$ be on the line $CD$ (may be outside of the segment $CD$). $BP$ intersects $\Gamma_2$ at $Q$. Prove that $A,D,P,Q$ lie on a circle. Solution:- After Inverting around $A$ with any arbitary radius, the problem becomes as ... Inverted Problem wrote: Let $A$ be any point in a plane and two parallel lines $\Gamma_1'$ and $\Gamma_2'$ are on the same side of $A$. Any arbitary line $l_1$ passing through $A$ intersects $\Gamma_1'$ at $C'$ and $\Gamma_2'$ at $B'$, another arbitary line $l_2$ passing through $A$ intersects $\Gamma_2'$ at a point $Q'$ and another arbitary line $l_3$ passing through $A$ intersects $\Gamma_1'$ at $D'$ and if $P'=\odot(AC'D')\cap\odot(AB'Q')$, then $\overline{P'-D'-Q'}$. Now we just need to chase angles for this..... Let $P'D'\cap l_2=(Q')'$ we will prove that $(Q')'\equiv Q'$. So, $$\angle P'D'C'=180^\circ-\angle B'AP'=\angle P'Q'B'=\angle P'(Q')'B'\implies (Q')'\equiv Q'.$$Now inverting back we get that $A,D,P,Q$ are concyclic. $\blacksquare$

Dear Mathlinkers, A is the Miquel's point of the rriangle BCD.... Sincerely Jean-Louis

Why homothety and inversion if you have an easy angle chasing solution? Draw the common tangent at $A$, and name it $XY$ with $\angle YAD$ acute ( It may work without this but I want to avoid configurational issues). $\angle YAD=\angle ACD =\angle BCP$. Again,$\angle XAQ=\angle QBA=\angle CBP$. Adding the above two conditions we get, $\angle YAD +\angle XAQ =\angle BCP +\angle CBP$. Hence,$180-\angle QAD=\angle QPD$. This, $A,D,P,Q$ are concyclic