Let $T$ be the tangency point and $O$ be the center of the hexagon. Then from $\angle PAB=150^\circ=\angle QDC$ and $$PT\cdot TQ=OT^2\quad\iff\quad\frac{PA}{AB}=\frac{CD}{DQ}$$we obtain that $\Delta PAB \sim\Delta QDC$. Hence the angle between $QC$ and $PB$ is equal to $$\angle APB+\angle CQD=\angle APB+\angle PBA=30^\circ.$$

We will prove that the answer is $30^\circ$. Consider the duality transform wrt circumcircle. Quote: Let $O$ be the center of circumcircle, $X$ be any point on the minor arc $EF$, and let $XA$ meets the tangent to circle at $B$ by point $R$, $XD$ meets the tangent at $C$ by point $S$. Need to prove that $\angle ROS=150^\circ.$ Easy count of angles shows that $\angle (DC;RB)=\angle (CS;BA)=\angle (SD;AR)=\frac{\pi}{2}$, so there exists a spiral similarity that maps $S$, $D$, $C$ onto $A$, $R$, $B$ respectively. Thus $|CS|\cdot |BR|=|AB|\cdot |CD|=|OC|\cdot |OB|$ means $\triangle OCS\sim \triangle RBO$ and $\angle COS+\angle ROB=90^\circ$. Finally $\angle ROS=90^\circ +\angle BOC=150^\circ$ as required.

Attachments: