Let $\angle ADB=\angle ACB=x$ and $\angle ACD=\angle ABD=y$. Note that $\angle BPC=90^\circ-x-y$ so $\angle APB=90+x+y$. Now since $A$ is the circumcenter of $\triangle DPK$ we get $\angle DAK=180^\circ-2x-2y=\angle DBC$. This gives $\angle BDC=x+y=\angle CAB$ (in particular $\triangle DBC$ is isosceles). We also have \[\cos(2x)=\frac{AP}{AH}=\frac{AK}{AH}=\frac{\sin(\angle ABK)}{\sin(\angle ABH)}\cdot\frac{BK}{HB}=\frac{\cos(2x+y)}{\cos(x+y)}\cdot\frac{1}{\cos(x)}\]This further yields \[\tan(x+y)\tan(x)-1=\cos(2x)\iff \tan(x+y)=\sin(2x)\]Finally, since $\angle PAK=2x$, this implies that \[\frac{PK}{AK}=\frac{HB}{AK}\iff PK=HB\]Therefore, $\triangle PKB$ is isosceles so $\angle KPB=\angle KBP=x$ and this concludes since $\angle APK=\angle AKP=\angle BKC=90^\circ-x$. $\blacksquare$

Nice problem, here's a sketch for a synthetic solution (Edit: a small mistake, it should be $BD=BC$ instead of $DB=DC$.)

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Another idea for trig bash is adding point F on AD s.t BF=BD=BC , this allows u to try to prove similarity of triangles which i feel like its easier to work with sines here than other trig bashes.

Claim 1: $BC = BD$ Let $\gamma := \angle BCD$ we have: \[ \angle BPC = 90-\gamma \implies \angle KAD = 180 - 2 \gamma \implies \angle CBD = 180 - 2\gamma \implies \angle BDC = \gamma \implies BC=BD \] Next reflect $K$ over $A$ to get $X$. We claim $BX = BC = BD$. Since we have $AX=AK$, $X$ must lie on $DPK$: \[ \angle KXD = \angle KPD = 90 - \gamma = \frac{1}{2}\angle CBD \]Thus $B$ is the circumcenter of $CXB$, since we already know $BC=BD$. Lastly we claim that $XPHB$ is cyclic. Let $\varphi := \angle APK = \angle AKP = \angle BKC$: \[ \angle BXH = \angle BCH = 90 - \varphi = \frac{1}{2} \angle KAP = \angle HXP \]Thus we have \[ \angle APH = \angle APK + \angle BPH = \varphi + \angle BCH = \varphi + 90 - \varphi = 90\]

Let \( Q \) be the reflection of \( P \) about \( AC \). Observe that \[ 180^\circ - \angle AQC = 180^\circ - \angle APC = \angle APD = \angle ADC. \]It follows that \( Q \) lies on \( (ABCD) \). Next, note that: \[ \angle BQK = \angle AQB - \angle AQK = 180^\circ - \angle ACB - \angle AKP = 180^\circ - \angle KCB - \angle BKC = 90^\circ. \]Thus, \( BHKQ \) is cyclic. To finish: \[ \angle APH = \angle APK + \angle KPH = \angle AKP + \angle KQH = \angle BKH + \angle KBH = 90^\circ. \]

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Let $\omega$ be the circle centered at $A$ through $D$. The key point is $E$, the other intersection of $\omega$ with $(ABCD)$. Claim: $KE=KP$. Proof: As $AE=AD$, $A$ is the arc midpoint of $DE$ in $(ABCD)$. Since $AK = AD = AE$, by Incenter-Excenter $K$ is either the incenter or $C$-excenter of $\triangle CDE$. In any case, $DK$ bisects (either internally or externally) $\angle EDP$. As $E,K,P,D$ all lie on $\omega$, this means that $KE=KP$. Hence, $E$ and $P$ are reflections about $CD$. Now, we let $EK$ intersect $(ABCD)$ at $X$. By Reim, $XC // KP$ so $XC \perp BC$ so $X$ is the antipode of $B$. So $\angle BEK = BEX = 90^{\circ}$, which means $BEKH$ is cyclic with diameter $BK$. Now $\angle EBH = \angle EKA$ and \[ \angle BEH = \angle BKH = \angle AKP = \angle EKA = \angle AEK\]($E$, $P$ are reflections about $CD$). Thus $\triangle EHB$ spirals to $\triangle EAK$, by swapping the spiral $\triangle BEK$ spirals to $\triangle HEA$ too so $\angle HEA = 90^{\circ}$. Reflecting over $CD$, $\angle HPA = 90^{\circ}$ as desired.

Let $S$ be the orthocenter of $\triangle ABC$. Claim: Point $P$ lies on $(SAC)$. Proof: Using directed angles, we have $\angle CPA = \angle DPA = \angle ADP = \angle ADC = \angle ABC = \angle CSA$. Claim: $(AKP)$ is tangent to $SA$. Similarly, $(CKP)$ is tangent to $SC$. Proof: We have $\angle SAK = \angle CBH = \angle BKH = \angle PKA = \angle APK$ so $(AKP)$ is tangent to $SA$. Now, $\angle CPK = \angle CPA - \angle KPA = 180^{\circ} - \angle ASC - \angle CAS = \angle SCA = \angle SCK$, so $(CKP)$ is tangent to $SC$ as well. At this point, we are going to generalize our problem. Note that the original problem is the special case of the following generalization when $P$, $K$ and $B$ are collinear. Generalization: Let $\triangle SAC$ be a triangle with orthocenter $B$. Let $H$ be the foot of the altitude from $S$, and let $K$ be the point on side $AC$ such that $BK \parallel SA$. Point $P$ lies on $(SAC)$ such that $(AKP)$ is tangent to $SA$ and $(CKP)$ is tangent to $SC$. Then we have $\angle APH = 90^{\circ}$. [asy][asy] size(13cm); import geometry; draw(unitcircle, blue+white); pair S = dir(50); pair A = dir(190); pair C = dir(320); pair B = orthocenter(S,A,C); pair H = intersectionpoint(line(S,B),line(A,C)); pair K = intersectionpoint(line(A,C),parallel(B,line(S,A))); transform reflect = reflect(perpendicular((0,0),line(A,C))); pair T = reflect * S; transform reflect1 = reflect(perpendicular((0,0),line(T,K))); pair P = reflect1 * T; transform reflect2 = reflect(line(A,C)); pair L = reflect2 * B; transform reflect3 = reflect(perpendicular((0,0),line(P,H))); pair Ap = reflect3 * P; dot(S); dot(A); dot(C); dot(B); dot(H); dot(K); dot(T); dot(P); dot(L); dot(Ap); label("$S$", S, dir(S)); label("$A$", A, dir(A)); label("$C$", C, dir(C)); label("$B$", B, dir(B)); label("$T$", T, dir(T)); label("$H$", H, dir(H)); label("$L$", L, dir(L)); label("$P$", P, dir(P)); label("$K$", K, dir(K)); label("$A'$", Ap, dir(Ap)); draw(A -- S -- C -- cycle); draw(S -- L); draw(B -- K); draw(T -- P); draw(P -- Ap); draw(A -- P); draw(A -- B -- C); markrightangle(H,P,A); draw(circumcircle(A,K,P), orange); draw(circumcircle(C,K,P), orange); draw(circumcircle(K,H,L), purple); [/asy][/asy] Let $PK$ hit $(SAC)$ for the second time at $T$. Note that $\angle TSA = \angle TPA = \angle KPA = \angle CAS$ so $ST \parallel AC$. Thus, we have $\angle PKH = \angle PTS = \angle PLS = \angle PLH$ so $K$, $L$, $H$ and $P$ are cyclic. Now let $PH$ hit $(SAC)$ again at $A'$. Then note that $$\angle A'CA = \angle A'CT + \angle TCA = \angle A'PT + \angle CAS = \angle HPK + \angle CAS = \angle HLK + \angle CAS = \angle KBL + \angle CAS = \angle ASH +\angle HAS = 90^{\circ}$$Hence, $\angle APH = 90^{\circ}$ as desired.