Two circles, two secants and a pair of parallel lines

Chase. $A,B,P,P'$ are concyclic (on $\Gamma_1$), so
$$\angle(P'P,P'B)=\angle(AP,AB).$$
$P,A,Q$ are collinear, so $\angle(AP,AB)=\angle(AQ,AB)$.
$A,B,Q,Q'$ are concyclic (on $\Gamma_2$), so
$$\angle(AQ,AB)=\angle(Q'Q,Q'B).$$
Altogether, $\angle(P'P,P'B)=\angle(Q'Q,Q'B)$. Since $P'B$ and $Q'B$ are the same line $\ell$, $\angle(PP',\ell)=\angle(QQ',\ell)$, so $PP'\parallel QQ'$. $\blacksquare$

Remark. If the second line is tangent, replace the inscribed angle by the tangent-chord angle.

Lesson. Directed angles mod $180^\circ$ remove every "inside or outside?" worry.