Be $Q$ the midpoint of $AM,$ $T$ the intersection between the perpendicular bisector of $AM$ and $AP,$ and $R$ the foot of the perpendicular drawn from $B$ to $AG.$ $QT\parallel PM$ and $Q$ is midpoint of $AM$ $\to$ $T$ is midpoint of $AP.$ and $C$ is midpoint of $AG$ $\to$ $AC=CG$ $\to$ $MC\parallel BG$ $\to$ $BG=AD$ (because $AM=MB$) $\to$ $ADBG$ is a parallelogram and $AG=GB=BD=DA$ $\to$ $\angle{ADP}=\angle{PDB}=\angle{BGD}=\angle{DGA}=\alpha$ and $\angle{BAG}=\angle{BAD}=\angle{ABG}=\angle{ABD}=90-\alpha$ as $AP$ and $PB$ are tangents to $S$ $\to$ $\angle{BAP}=\angle{ABP}=2\alpha.$ $\angle{ABR}=90-\angle{BAR}=90-(90-\alpha)=\alpha$ Be $X$ the intersection point of $PM$ and $S$ $\angle{ABX}=\angle{ADX}=\alpha$ then $R,$ $X,$ and $B$ are colineal. $AMXR$ is cyclic, so $\angle{AMR}=\angle{AXR}=\angle{ADB}=2\alpha$ be $T1$ intersection between $MR$ and $AP$ $\to$ $AMT1$ is isoceles $\to$ $T1$ lies on the perpendicular bisector of $AM$ $\to$ $T1=T.$ Also, $\angle{MRA}=\angle{MXA}=90-\angle{MAX}=90-\alpha$ $\to$ $MR=AM.$ $QCRB$ is cyclic $\to$ $\angle{BQR}=\angle{BCR}=2\alpha$ $\to$ $QR\parallel AT$ ($\angle{BQR}=\angle{BAT}=2\alpha$) $\to$ as $QM=QA,$ $MR=RT$ $\to$ $AT=AB.$ Be $T'$ the point where $BG$ touches $AP$ by a little angle chasing we get: $\angle{ABT'}=\angle{AT'B}=90-\alpha$ $\to$ $T'=T.$ Then $G$ is the intersection of $PM$ and $BT$ ($M$ and $T$ being midpoints of $AB$ and $AP$ respectively) so it's the gravicenter.

Aldo Pacchiano wrote: Be $ T'$ the point where $ BG$ touches $ AP$ by a little angle chasing we get: $ \angle{ABT'} = \angle{AT'B} = 90 - \alpha$ $ \to$ $ T' = T.$ I can't see how this implies $ T'=T$ (we still don't know $ \angle{ATB}$, do we?)

Let $ I$ be the second intersection of $ PD$ with $ \mathcal{S}.$ Since $ \angle PBI = \angle IDB = \angle IBM,$ it follows that $ I$ is the incenter of $ \triangle PAB$ $ \Longrightarrow$ $ BD$ is the external bisector of $ \angle PBA$ $\Longrightarrow$ Parallel $ AC$ to $ BD$ cuts $ PB$ at $ N$ such that $ \triangle BNA$ is isosceles with apex $ B.$ Let $ \angle ABD = \theta.$ Since $ BNCM$ and $ ABIC$ are both cyclic quadrilaterals, we have $ \angle ACB = \angle AIB = 2\theta$ and $ \angle BCM = \angle BNM = \angle ACB - \angle ACM = 2\theta - (\pi - 2\theta) = 4\theta - \pi.$ On the other hand, if $ O$ is the center of $ \mathcal{S},$ we obtain $ \angle PAB = \angle AOI =\pi - 2\theta \Longrightarrow \angle APB = \pi - 2(\pi - 2\theta) = 4\theta - \pi = \angle BNM$ Thus, $ NM \parallel PA$ $\Longrightarrow$ $ N$ is the midpoint of segment $ PB,$ which implies that $ G \equiv PM \cap AC$ is the centroid of $ \triangle PAB.$