Let $E$ be the second common point of $DA$ and the circle. One can show that $A$ is between $D$ and $E$. Denote $\angle ELG = \alpha$, $\angle GLC = \beta$. Since $GLBC$ is a cyclic quadrilateral by hypothesis, we have $\angle GBC = \angle GLC = \beta$; since $ELGA$ is also cyclic, $\angle DAG = \alpha$. Thus $\alpha$ and $\beta$ are the measures of adjacent angles in a parallelogram, hence $\alpha+\beta = 180^{\circ}$. It follows that $E$, $L$ and $C$ are collinear.

Let $\angle LEA = \theta$, then $\angle LGB = \theta$ as $ELGA$ is cyclic. Hence $\angle LGB = \angle LDC = \theta$ due to $AB \parallel CD$.
Triangles $EDC$ and $DLC$ are similar as they share an angle at $C$ and $\angle CED = \angle CDL = \theta$. The similitude gives $\frac{CD}{CL} = \frac{CE}{CD}$, or $CD^2 = CL \cdot CE$. On the other hand, by power of a point, $CP^2 = CL \cdot CE$. Hence $CD = CP$. Since $CD = AB$ (opposite sides of a parallelogram), we obtain $AB = CP$, as desired.