Looks very intimidating. Let $H $ be the orthocenter of $\triangle AUV$ and let $\overline{UT} $ intersect $(PVT) $ again at $X $. By the bisector condition and $[UA] = [UM] \Rightarrow AUMT$ - cyclic $\Rightarrow \measuredangle UTA = \measuredangle UMA = \measuredangle MAU = \measuredangle CAV = \measuredangle HAU = \measuredangle UVH$. Since $PXVT $ - cyclic $\Rightarrow \measuredangle XPU = \measuredangle UTA = \measuredangle UVH$. Now a similar triangle chase. $\triangle PXU \sim \triangle TVU \sim \triangle AHU$ and $\triangle AHV \sim \triangle AUP$. $[PX]:[PU] = [AH]:[AU] = [VH]:[PU] \Rightarrow [PX] = [VH]$, combining this with $\measuredangle XPU = \measuredangle UVH \Rightarrow PXHV$ - isosceles trapezoid $\Rightarrow PXHV$ - cyclic $\Rightarrow PHVT$ - cyclic. Similarly we can obtain $NHUS $- cyclic, thus $(NUS) $ and $(PVT) $ meet on $\overline{AM} $.