Let $K$ be second intersection point of $BD$ with Г'. then $\cup AD= \cup AK$ so $\measuredangle{ABD}=\measuredangle{ACD}$ and $CD=BD=OO'$.

Sorry, but I don't see how exactly it was concluded that $\angle DBA = \angle DCA$.. Could someone explain this?

This is clearly a problem on translation... Let $OO' = d$. Translate the circle $\Gamma'$ through the vector $O'O$. This is just moving the circle $\Gamma'$ in the direction of $O'O$ and a distance $d$ such that it coincides with circle $\Gamma$. Suppose the translation sends $C$ to $C'$ and $A$ to $A'$. Clearly, because of the way it was constructed, $D$ is sent to $B$. Clearly, we have $DC= BC'$. Also $AA' = CC' = d$. Since $ACC'A'$ is a paralelogram, we have $A'C' \parallel AC$. Hence $A'C' \parallel AB$. Therefore, $BAC'A'$ is an isosceles trapezium, which means $BC' = AA'$. In conclusion, $CD = BC' = AA' = d$, which is a constant.

Let $E$ the second point of intersection of $\Gamma$ and $\Gamma'$. If we consider the homothecy that sends $\Gamma$ to $\Gamma'$, it is clear that sends $B$ to $D$ and $A$ to $E$, now notice that this implies that sends $BA \cap \Gamma' = C$ to a point in $\Gamma$ that sattisfies $DE \cap \Gamma = F$, then it's clear that $BA\parallel DE$ hence $AC \parallel DE$ which directly implies that $CD=AE$ which is clearly constant

Define: $OO' \cap \Gamma '=H $, Let $BO \cap \Gamma =G$ and Let $BD \cap \Gamma =E$, then we have, $$\angle EBD=\angle ABP-\angle EBO=\frac{1}{2} \angle AOG-\angle EBO=\frac{1}{2} \angle AOO'-\frac{1}{2} \angle EBO=\frac{1}{2} \angle AO'O-\angle DCH=\angle ACH-\angle DCH=\angle ACD$$Hence, $CD=BD=OO' \longrightarrow$ which is obviously fixed no matter where $B$ lies!