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$D$ is the circumcenter of $BAC$.
$\therefore \measuredangle BAD=\measuredangle DBA$.

$M$ is the circumcenter of $XAY$.
$\therefore \measuredangle XAM=\measuredangle MXA$.

$M$ is the circumcenter of $DTP$.
$\therefore \measuredangle MDT=\measuredangle DTM$.

Now, let's show that $ADMT$ cyclic.
\[
\begin{equation}
\begin{split}
\measuredangle DAM &=\measuredangle XAM-\measuredangle BAD \\
&=\measuredangle MXA-\measuredangle DBA \\
&=\measuredangle DXB+\measuredangle XBD \\
&=\measuredangle XDB=\measuredangle MDT=\measuredangle DTM
\end{split}
\end{equation}
\]
$\therefore ADMT$ cyclic.
\[
\begin{equation}
\begin{split}
\therefore \measuredangle DAM&=\measuredangle DTM \\
&=\measuredangle MDT=\measuredangle MAT
\end{split}
\end{equation}
\]
$\therefore AM$ bisects $\angle TAD$