Super hard Euclidean Geometry

\begin{align*} \angle TDM &= \angle YDC \\ &= \angle DYA - \angle DCY \text{ (exterior angle = sum of opposite interior angles in triangle }DCY) \\ &= \angle MAY - \angle DCY \text{ ($M$ is centre of circle through $XAY$, so $\angle MYA = \angle MAY$)} \\ &= \angle MAY - \angle DAY \text{ ($D$ is centre of circle through $BAC$, so $\angle DCY = \angle DAY$)} \\ &= \angle MAD \end{align*}

But $\angle TDM = \angle MTD$ (because $M$ is centre of circle through $PTD$, so $MD = MT$). Thus $\angle MTD = \angle MAD$, and so $MTAD$ is a cyclic quadrilateral. And $MD = MT$. Hence $\angle TAM = \angle MAD$.

QED