Take $G=CN\cap AB$ and $H$ the reflection of $C$ in line $BN$. Claim. $B,K,N,G,H$ lie on a circle tangent to $TN$. Proof: It's clear that $H\in (BNK)$, so it's enough to prove that $G\in (BNK)$ and that $TN$ is tangent to $(BNKG).$ To prove that $G\in (BNK)$ just note that $\angle{BGN}+\angle{BKN}=\angle{BCN}+\angle{BKN}=180^0$, since $BG=BC$ and $K$ is the orthocenter of $\Delta{BNC}.$ Moreover, note that \begin{align*} \angle{TNG} & =\angle{GNA}-\angle{TNA} \\ & =\angle{DNC}-\angle{MDN} \\ & =\angle{BNC}-\angle{BCN} \\ & =180^0-2\angle{BCN}-\angle{NBC}\\ & =\angle{ABC}-\angle{NBC}\\ & =\angle{GBN} \end{align*}so $TN$ is tangent to $(BKNG)$. $\blacksquare$ Consequence: $H\in TB$ Proof: Apply Pascal on $HKNNGB$ and it's clear. This obviously is enough to end the problem.

Redefine $T$ as the point so that $TN \parallel DM$ and $\angle NBC=\angle TBN.$ Then let $J$ be the reflection of $C$ with respect to $M$ and let $H$ be on side $AB$ such that $BKNHJ$ is cyclic. So by angle chasing $C,N,H$ are collinear and $TN$ is tangent to the circumcircle of $NKB$. By Pascal's theorem on $NNKJBH$, $C, A, T$ are collinear and the result follows.

We firstly define $C'$ as reflection of $C$ over $M$ and $T$ as $CN \cap AB$. $KMBD$ and $MNCD$ is cyclic Now, $\angle BC'K=\angle MCB=90- \angle MBD=\angle BND$, giving $KNBC'$ cyclic. Then it follows $K$ is orthocentre of $\triangle BNC$. As $BK$ bisects $\angle FBC$,it follows the triangle is isosceles. Now, $\angle DKB=\angle BMD=\angle NCD=\angle BFN$, giving $BFNK$ cyclic. Thus,$BC'FNK$ is cyclic. Now, $\angle TNF=\angle ANF-\angle ANT=\angle FBK-\angle MDK=\angle FBK-\angle NBK=\angle FBN$ Thus, $TN$ is tangent to $(BC'FNK)$. Now, Pascal on $FBC'KNN$ gives $B--C'--T$ Now, add in the ,midpoint of $BC$ and this finishes.

It is clear that $K$ is the orthocenter of $\Delta KBC$. Define $F=BA\cap KC$. Thus, $\angle FBK=\angle ABK = \angle DBK = \angle DNC = \angle KNF$. $\newline$ $\Longrightarrow F\in (NBK)$. $\newline$ Moreover by parallel lines, we have: $\angle KNT = \angle KDN =\angle KBM = \angle KBN$ $\newline$ $\Longrightarrow$ $NT$ is tangent to $(NBK)$ $\newline$ Define $E=CM\cap (NBK)$. It is well-known that $E$ is the reflection of $C$ over $NB$. $\newline$ Now, Pascal on $NNKEBF$ gives $EB\cap NN = T'\in AC$ $\newline$ $\Longrightarrow T=T'$ Finally, we get: $\newline$ $\angle TBC = \pi-\angle CBE = \pi-2\angle CBN$. $\newline$ Which proves that $BM$ is the external angle bisector of $\widehat{TBC}$ $\newline$ $\blacksquare$