Let $M,N,K$ be midpoints of $AB,CD,CQ$ and $T$ be an arbitrary point on $PL$. Let $N'$ be reflection of $N$ w.r.t $PL$. Claim $:PLKMA$ is cyclic. Proof $:$ Note that $PQ.QC=QA.QD \implies PQ.QK=QA.QL$ so $PLKA$ is cyclic. Also $\angle LKM = \angle DNM = 180 - \angle A$ so $LKMA$ is cyclic. Now note that $N'L = NL = CK$ and $AK = LM$ and $\angle N'LM = 180 - \angle ALM + \angle N'LD = 180 - \angle ALM + \angle NLT - \angle DLT = 180 - \angle KPL + \angle NLD - 2\angle TLD = 180 - \angle ALP = \angle AKC$ so $N'LM$ and $AKC$ are congruent so $AC = N'M$ and $TN+TM = TN'+TM > N'M = AC$ as wanted.