Let $r$ be the radius of the circle. By the law of sines $\tfrac{|MB|}{\angle BVM} = \tfrac{r}{\sin \angle MBV}$. So $\angle BVM$ is maximized exactly if $\angle MBV = 90^{\circ}$.

Kezer wrote: Let $r$ be the radius of the circle. By the law of sines $\tfrac{|MB|}{\angle BVM} = \tfrac{r}{\sin \angle MBV}$. So $\angle BVM$ is maximized exactly if $\angle MBV = 90^{\circ}$. Not to be picky, but in the actual contest you would have lost one point for this solution because of not considering the special case $\sin(...)=0$...

Thanks Tintarn, true. I remember that I admittedly forgot to consider such cases in math olympiads quite often as well back then...

Suppose line $VB$ intersects $k$ another time at $G$ and line $VM$ intersects at $F$. By Thales' Theorem $\angle{FGV}=90$. Since $VF$ remains constant all time, we want to minimize $VG$ and maximize $GF$, thus, when $VG$ is perpendicular to $MB$ it is minimal and $FG$ is maximal $QED$

Let the $\gamma$ be one of the two circles through $M$ and $B$ tangent to $k$ at $T$. We claim that $T$ is the desired point. Clearly, except the symmtric case for the second tangency circle, $\angle BVM<\angle BTM$ as $MB$ is a chord of $\gamma$, and $V$ lies outside of $\gamma$ for all $V\neq T$.