Solution:

(1) The triangles $A C D$ and $A E F$ are directly congruent: We work with directed angles modulo $\pi$. We have $\text{A D C = A D B = A F B = A F E}$ and $\text{D C A = B C A = B E A = F E A}$ (for the middle equality sign the inscribed angle theorem was used in both cases). Since, moreover, $|C D| = |E F|$ holds by assumption, the claim follows by the ASA congruence theorem.

(2) $A$ lies on the same side of the line $C D$ as $F$: The overlap region of the two circles contains the segment $B E$ and meets the line $C D$ only at the point $B$.

(3) The triangles $C D Q$ and $E F P$ are directly congruent: As in (1), $\text{Q D C = Q D B = Q F B = P F B}$ etc.

(4) There exists a rotation about $A$ that maps the points $C, D, Q$ to the points $E, F, P$: this follows from (1) and (3).

(5) $A$ lies on the same side of the line $B F$ as $C$: From (2) it follows that $A$ and $Q$ lie on the same side of the line $C D$, and together with (4) it follows that $A$ and $P$ lie on the same side of the line $E F = B F$, hence also $A$ and $C$.

(6) $B A$ is the bisector of the interior angle of the triangle $B F C$ at $B$: By (1) the distances from $A$ to the extended sides are equal, and by (2) and (5) it is an internal bisector.

(7) $C M$ is the bisector of the interior angle of the triangle $B F C$ at $C$: This follows from the inscribed angle theorem.

(8) Analogously, $F N$ is the bisector of the interior angle of the triangle $B F C$ at $F$. Let $I$ be the incenter of the triangle $B F C$.

(9) $M$ lies on the arc $B A$ of $\Gamma$ not containing $C$: From (2) and (5) it follows that the points on $\Gamma$ have the cyclic order $B, E, A, C$ resp. $B, P, C$ (equally oriented). The directed arcs $E A$ and $A C$ are equal by (4), so $\frac{1}{2} B C < B A$ and hence $B M = \frac{1}{2} B P < \frac{1}{2} B C < B A$.

(10) $N$ lies on the arc $A B$ of $\Gamma'$ not containing $F$: analogous to (9).

(11) $I$ is an interior point of $C M$ and $F N$: $I$ lies on the line $A B$, hence the claim follows from (9) resp. (10).

(12) $I$ is an interior point of $B A$: follows from (9) or (10).

(13) We have $|C I| \cdot |I M| = |F I| \cdot |I N|$: By the power of a point (chord theorem) in $\Gamma$, $|C I| \cdot |I M| = |B I| \cdot |I A|$, and by the power of a point (chord theorem) in $\Gamma'$, $|B I| \cdot |I A| = |F I| \cdot |I N|$.

(14) By the converse of the power of a point (chord theorem), it follows with (11) and (13) that $CNMF$ is a cyclic quadrilateral.