Claim: $B,N,P,D,Q,S$ are concyclic Proof. Since, $B$ lies on $(ACD)$ and $BN$ and $BM$ are perpendicular to $AD$ and $AC$ respectively, we get that $MN$ is the Simson Line of $B$ w.r.t. $\triangle ACD$ and hence, $BQ \perp CD$. Similarly, we get that $DS \perp BC$. $\square$ Also, note that quadrilaterals $AMBN$ and $APDT$ are cyclic. Now, for part a), $\angle AMN = \angle ABN = \angle PBN = \angle PSN = \angle PQN = \angle PDN = \angle PDA = \angle PTA \implies PQ \parallel AC \parallel NS$. For part b), Since we have gotten that $PQ \parallel NS$ and also the four points $P,Q,S,N$ lie on a circle $\implies \square PQSN$ is an isosceles trapezium from which the result follows. For part c), if $AC$ is diameter of $(ABCD)$, then $\angle ABC = 90^{\circ} = \angle ABS = \angle PBS = \angle PNS \implies PNSQ$ is a rectangle. Only if part of c) can be easily proved working backwards above. And we are done.

I will only prove $NP = SQ$, the other parts are pretty much the same. Clearly $AMBN$ is cyclic (with two right angles), so $\angle DQN = 180^{\circ} - \angle QDN - \angle DNQ = \angle ABC - \angle ANM = \angle ABC - \angle ABM = \angle CBM$, hence $\angle CBM + \angle CQM = 180^{\circ}$ and $BCQM$ is cyclic. Thus $\angle BQC = \angle BMC = 90^{\circ}$. Analogously $\angle CSD = 90^{\circ}$. Therefore the hexagon $BNPDQS$ is cyclic (with diameter $BD$); in particular $NPQS$ is cyclic. Now we have $\angle PQN = \angle PDN = \angle BAD - 90^{\circ} = 90^{\circ} - \angle BCD$ and $\angle QPS = \angle QBS = 90^{\circ} - \angle BCD$. Therefore $\angle PQN = \angle QPS$, concluding $NP = SQ$.