Let $LD$ intersect $(PBK)$ again at $E$. Claim: $EMDK$ is cyclic. Proof: We have $$\angle KED=\angle KEL=\angle KBL=\angle KMD.$$ Claim: $E$, $M$ and $P$ are collinear. Proof: Note that $$\angle MKD=\angle BKD=\angle BAD=\angle PBC=\angle PBL=\angle PKL$$and $$\angle KMD=\angle KBL=\angle KPL$$Hence, $\bigtriangleup MKD\sim \bigtriangleup PKL$. This means that $\angle KDM=\angle KLP$, or $$\angle KEM=180^{\circ}-\angle KDM=180^{\circ}-\angle KLP=\angle KEP$$ Now let's finish. Note that $$\angle DCL+\angle CLD=\angle PCB+\angle BPE=\angle PBC+\angle BPE=90^{\circ}.$$We're done.

Let $N$ be the midpoint of $BC$, $E=PA\cap DN$ and $T=DN\cap BM$. Claim: $TP\parallel BC$. Proof: First define $T'$ to be the intersection of parallel to $BC$ through $P$ with $ND$ we show $T=T'$. We know $M$ is a midpoint so $$-1=(A,D;M,P_{\infty})\overset{B}{=}(A,D;K,C)$$Now let $K'=T'M\cap (ABC)$ then $$-1=(B,C;N,P_{\infty})\overset{P}{=}(T',N;E,D)\overset{B}{=}(A,D;K',C)$$So $K=K'$ and therefore $T=T'$ meaning $TP\parallel BC \qquad \square$. Now it's just angle chasing, note that this means $TPDK$ is cyclic so $$\angle NDP=\angle TDP=\angle TKP=\angle BKP=\angle BLP=\angle NLP$$That means $PNDL$ is cyclic, but $\angle PNL=90^{\circ}$ so $\angle PDL=90^{\circ}$

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Can someone provide a solution using Apollonius' Lemma?