Let $ABCD$ be the base of this pyramid and $ E$ its opposite vertex. Let $ X,Y,Z,W$ be the tangency points of its insphere $\omega$ with the faces $ EAB,$ $EBC,$ $ECD,$ $EDA$ and $ P$ its tangency point with the base $ABCD.$ It is easy to figure out that $ \triangle EXB \cong \triangle EYB,$ because of $ BX = BY$ and $ EX = EY.$ Mutatis mutandis $ \triangle EYC \cong \triangle EZC,$ $\triangle EZC \cong \triangle EWD$ and $\triangle EWA \cong \triangle EXA.$ Then we get
$ \angle CYB = 360^{\circ} - (\angle EZC + \angle EXB)$
$ \angle AWD = 360^{\circ} - (\angle EZD + \angle EXA)$
$ \angle CYB + \angle AWD = 720^{\circ} - (\angle EZC + \angle EXB + \angle EZD + \angle EXA)$
$\angle CYB + \angle AWD = \angle CZD + \angle AXB$
Since $ AX = AW = AP,$ $BX = BY = BP,$ $CY = CZ = CP$ and $DZ = DW = DP$ we get
$ \angle AXB = \angle APB , \ \angle BPC = \angle BYC$
$\angle CZD = \angle CPD, \ \angle AWD = \angle APD$
$\Longrightarrow \angle APD + \angle BPC = \angle CPD + \angle APB$
$ \Longrightarrow \angle APD + \angle BPC = 180^{\circ} , \ (*)$
Let $ M,N,L,K$ be the orthogonal projections of $ P$ onto the edges $ AB,BC,CD,DA.$ Since the quadrilaterals $ PMBN,$ $PNCL,$ $PLDK,$ $PKAM$ are all cyclic, it follows that
$ \angle APD = \angle KMA + \angle KLD , \ \angle BPC = \angle NMB + \angle NLC$
Combining with $ (*)$ yields $ \angle KLM + \angle NMK = 180^{\circ}$ $\Longrightarrow$ $ MNLK$ is cyclic.