Invert about $A$ with radius $AB$. Then, we see that $C,P$ and $D,Q$ are inverses, as $(ABC) \rightarrow BB$. Then, we know that $\angle{ACD}=\angle{AQP}$ from which we have $\angle{PCD}+\angle{DQP}=180$, implying the desired conclusion

From tangency we have $ \measuredangle ACB = \measuredangle ADB = 90^{\circ} = \measuredangle ABP = \measuredangle ABQ $, thus $\Delta ACB \sim ABP $ and $ \Delta ADB \sim \Delta ABQ $ $$ AC . AP =AB^2 = AD . AQ $$Hence, we get $C,P,D,Q$ concylic, as desired.

Note that triangle $ABP$ is right-angled with $\angle ABP=90^{\circ}$ and $BP$ is its altitude. Hence, $AC\cdot AP=AB^2$. Similarly, $AD\cdot AQ=AB^2$. Thus, $AC\cdot AP=AD\cdot AQ$, so $PCDQ$ is cyclic, as desired.

$\angle DQB=90^\circ-\angle DBQ=\angle ABD=\angle ACD$. Done.

BAC=1/2(BC) --> CPQ=90-BAC= 90 - 1/2 (180 - AC) = AC/2. ADC= AC/2, so CDQ = 180-AC/2. Therefore, CPQ+CDQ=180. Am I wrong about this, because this seems kind of easy?