$\triangle PAD$ and $\triangle PBC$ are similar with corresponding medians $PN$ and $PL$ $\Longrightarrow$ $\triangle PND$ and $\triangle PLC$ are similar $\Longrightarrow$ $\angle DPN=\angle CPL.$ But from $NK \parallel BD$ and $LK \parallel AC,$ we have $\angle DPN=\angle KNP,$ $\angle CPL=\angle KLP$ $\Longrightarrow$ $\angle KNP=\angle KLP$ $\Longrightarrow$ $\odot(PKL) \cong \odot(PNK).$ By similar reasoning, we obtain $\odot(PKL) \cong \odot(PLM),$ $\odot(PLM) \cong \odot(PMN)$ and the conclusion follows.