Extend $AP$ and $DP$ till they meet $DC$ and $AB$ at $X$ and $Y$ respectively. Then $DAYX$, $DAYL$, $DXLA$ are all cyclic due to angle chasing. Hence, $DAYLX$ is cyclic and $L$ is the midpoint of the small arc $XY$. Moreover, $XY$ is antiparallel to $DA$ which is antiparallel to $BC$. Thus, $XY$ is parallel to $BC$. It follows that $K$ is the midpoint of $XY$. However, $KL \perp XY \rightarrow KL \perp BC$.

Let $AP, DP$ meet $CD, AB$ at $E, F$. $ADEF$ - cyclic with diameter $EF$. By Reim's $EF \parallel BC$. By homotheties at $A$ and $D$, $K$ - midpoint of $EF$. Since $L$ - midpoint of the arc $EF$, $KL \perp EF$.

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A solution from a different perspective: First of all, $\angle{KAD}=\angle{DNP}+\angle{ADP}=(90^\circ-\angle{C})+(90^\circ-\angle{B})=180^\circ-\angle{C}-\angle{B}$. Similarly, $\angle{KDA}= 180^\circ-\angle{C}-\angle{B}$, so $\angle{KAD}=\angle{KDA}$. From Ceva's Theorem in trigonometric form in $\triangle{LDA}$, we get: $$\frac{sin(\angle{KLA})}{sin(\angle{KAL})}*\frac{sin(\angle{KAD})}{sin(\angle{KDA})}*\frac{sin(\angle{KDL})}{sin(\angle{KLD})}=1 \Longleftrightarrow$$$$\frac{sin(\angle{KLA})}{sin(\angle{KLD})}*\frac{sin(\angle{KDL})}{sin(\angle{KAL})}=1 \Longleftrightarrow$$$$\frac{sin(\angle{KLA})}{sin(\angle{KLD})}=\frac{sin(\angle{KAL})}{sin(\angle{KDL})}=\frac{sin(\angle{B}-45^\circ)}{sin(\angle{C}-45^\circ)}$$Since $\angle{KLA}+\angle{KLD}=\angle{B}+\angle{C}-90^\circ<90^\circ$ and all angles are acute, we conclude that $\angle{KLA}=\angle{B}-45^\circ$ and that $\angle{KLD}= \angle{C}-45^\circ$. Now, $\angle{KLA}+\angle{CEL}=(\angle{B}-45^\circ)+(135^\circ-\angle{B})=90^\circ$, which ends the proof. Note: The angle computations have been done according to the figure below. The cases that $\angle{B}<45^\circ$ or $\angle{C}<45^\circ$, which imply that $K$ lies outside of $\triangle{LDA}$ are analogous.

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Observe by some angle chase thag $K$ is the circumcenter of $\triangle ADL$ now take another points $B^*, C^*$ on circle $(ADL)$ such that $K \in B^*C^*$ with circumcenter $K$ now observe as $DK=DB^*=DC^*$ with $\angle PDC=90^\circ$ so $P\in B^*D, C\in C^*D$ Now also as $\angle DB^*K=\angle DPC$ So $B^*C^*||BC$ and as $DB^*C^*L$ is cyclic With $\angle B^*KL=90^\circ$ hence $KL\perp BC$.

Really straightforward. We start off with making the following preliminary observations. Claim : Line $\overline{BC}$ is tangent to $(APD)$. Proof : This is easy to see, \[\measuredangle PDA = \measuredangle CDA + \measuredangle PDC = \frac{\pi}{2} + \measuredangle CBA = \frac{\pi}{2} + \measuredangle PBA = \measuredangle BPA \]which implies the claim. Now, we can also note that similarly, $\measuredangle DPC = \measuredangle DAP$ as well. This will be essential in future angle chases. Claim : $K$ is the circumcenter of $\triangle ALD$. Proof : We first check that $K$ is equidistant to points $A$ and $D$. Simply note that, \[\measuredangle DAK = \measuredangle DAP + \measuredangle PAK = \measuredangle DPC + \measuredangle BPA\]and \[\measuredangle KDA = \measuredangle KDP + \measuredangle PDA = \measuredangle DPC + \measuredangle BPA\]which implies that $\triangle KAD$ is isosceles and indeed $AK=KD$. Further, \begin{align*} \measuredangle AKD & = 2(\measuredangle KAD + \measuredangle PDK)\\ &= 2(\measuredangle APB + \measuredangle CPD)\\ & = 2(\measuredangle PAD + \measuredangle ADP)\\ & = 2(\frac{\pi}{4} + \measuredangle PAD + \frac{\pi}{4} + \measuredangle ADP)\\ &= 2(\measuredangle LAD + \measuredangle ADL)\\ &= 2\measuredangle ALD \end{align*}which finishes the proof of the claim. Now we are almost there, simply angle chase to victory. Let $R = \overline{DL} \cap \overline{BC}$ and $S = \overline{LK} \cap \overline{BC}$. Then, \begin{align*} \measuredangle RSL & = \measuredangle SRL + \measuredangle RLS\\ &= \measuredangle CRD + \measuredangle RLK\\ &= (\frac{\pi}{4} + \measuredangle CPD) + \measuredangle KDL\\ &= (\frac{\pi}{4}+ \measuredangle CPD) + (\frac{\pi}{4} + \measuredangle DPC)\\ &= \frac{\pi}{2} \end{align*}which finishes the proof.