C

$\because$在$BD$的同侧作等腰$Rt\triangle ABC$和等腰$Rt\triangle ADE$，$\angle ABC=\angle AED=90^{\circ}$，
$\therefore \angle BAC=45^{\circ}$，$\angle EAD=45^{\circ}$，
$\therefore \angle CAE=180^{\circ}-45^{\circ}-45^{\circ}=90^{\circ}$，
即$\angle CAM=\angle DEM=90^{\circ}$，
$\because \angle CMA=\angle DME$，
$\therefore \triangle CAM$∽$\triangle DEM$，故①正确；
由已知：$AC=\sqrt{2}AB$，$AD=\sqrt{2}AE$，
$\therefore \frac{AC}{AB}=\frac{AD}{AE}$，
$\because \angle BAC=\angle EAD$
$\therefore \angle BAE=\angle CAD$
$\therefore \triangle BAE$∽$\triangle CAD$，
$\therefore \frac{BA}{AC}=\frac{BE}{CD}$，
即$\frac{BE}{CD}=\frac{BA}{\sqrt{2}BA}$，
即$CD=\sqrt{2}BE$，故②错误；
$\because \triangle BAE$∽$\triangle CAD$
$\therefore \angle BEA=\angle CDA$
$\because \angle PME=\angle AMD$
$\therefore \triangle PME$∽$\triangle AMD$
$\therefore \frac{MP}{MA}=\frac{ME}{MD}$，
$\therefore MP\cdot MD=MA\cdot ME$，故③正确；
由②$MP\cdot MD=MA\cdot ME$
$\angle PMA=\angle DME$
$\therefore \triangle PMA$∽$\triangle EMD$
$\therefore \angle APD=\angle AED=90^{\circ}$
$\because \angle CAE=180^{\circ}-\angle BAC-\angle EAD=90^{\circ}$
$\therefore \triangle CAP$∽$\triangle CMA$
$\therefore AC^{2}=CP\cdot CM$
$\because AC=\sqrt{2}AB$，
$\therefore 2CB^{2}=CP\cdot CM$，故④正确；
即正确的为：①③④，
故选：$C$.