B

如图，过点$A$作$AP\bot OD$交$OD$的延长线于点$P$，设$MN$交$OD$一点$T$.

$\because \angle AOB=90^{\circ}$，$HN\bot OD$，
$\therefore \angle TON+\angle ONT=90^{\circ}$，$\angle H+\angle ONT=90^{\circ}$，
$\therefore \angle NOT=\angle H$，
$\because BM=BH$，
$\therefore \angle H=\angle BMH$，
$\because AP\bot OP$，$NH\bot OP$，
$\therefore AP$∥$NH$，
$\therefore \angle DAP=\angle BMH$，
$\therefore \angle NOT=\angle DAP$，
在$\triangle OTN$和$\triangle APD$中，
$\left\{\begin{array}{l}{∠OTN=∠P=90°}\\{∠NOT=∠DAP}\\{ON=AD}\end{array}\right.$，
$\therefore \triangle OTP$≌$\triangle APD\left(AAS\right)$，
$\therefore OT=AP$，
在$\triangle APO$和$\triangle OTH$中，
$\left\{\begin{array}{l}{∠AOP=∠H}\\{∠P=∠OTH=90°}\\{AP=OT}\end{array}\right.$，
$\therefore \triangle APO$≌$\triangle OTH\left(AAS\right)$，
$\therefore OA=OH=12$，
$\because \angle MDT+\angle DMT=90^{\circ}$，$\angle BOD+\angle AOP=90^{\circ}$，$\angle DMT=\angle NOT$，
$\therefore \angle BDO=\angle BOD$，
$\therefore BD=OB$，
$\therefore 5+BM=12-BH$，
$\therefore 5+BM=12-BM$，
$\therefore BM=3.5$.
故选：$B$.