Assume $ AC \perp BD$ and let $ E \equiv AC \cap BD$ be the diagonal intersection. Let $ F$ be intersection, other than $ E,$ of circumcircles $ \odot(ABE), \odot(CDE)$ with diameters $ AB, CD.$ Let $ P$ be intersection, other than $ F,$ of circumcircles $ \odot(FBC), \odot(FDA).$ Then $ \angle FCA = \angle FCE = \angle FDE = \angle FDB$ and $ \angle FAC = \angle FAE = \angle FBE = \angle FBD$ $ \Longrightarrow$ $ \triangle FAC \sim \triangle FBD$ are similar $ \Longrightarrow$ right $ \triangle FAB \sim \triangle FCD$ are similar $ \Longrightarrow$ $ \angle FAB + \angle FDC = \angle FBA + \angle FCD = 90^\circ$ $ \Longrightarrow$ $ \angle PAB + \angle PDC = \angle PBA + \angle PCD = 90^\circ.$ $ \angle PBC + \angle PAD = \angle PFC + \angle PFD = \angle CFD = 90^\circ$ and $ \angle PCB + \angle PDA = \angle PFB + \angle PFA = \angle AFB = 90^\circ$ $ \Longrightarrow$ $ P$ is the desired point. $ \square$ ____________________________________________ Assume a point $ P$ with the desired properties exists. Let perpendiculars $ a, b, c, d$ to $ PA, PB, PC, PD$ at $ A, B, C, D$ pairwise intersect at $ X \equiv d \cap a, Y \equiv a \cap b,\ Z \equiv b \cap c, T \equiv c \cap d.$ Cyclic quadrilaterals $ PDXA \sim ZCPB$ are similar $ \Longrightarrow$ $ XYZT$ is cyclic, let $ (O)$ be its circumcircle. Let $ U, V$ be poles of $ XZ, YT$ WRT $ (O).$ Then $ \angle ZUX = 180^\circ - 2 \angle XYZ$ and from similar right $ \triangle PDX \sim \triangle ZCP,$ $ \angle XPZ = \angle DPC + 90^\circ = \angle XYZ + 90^\circ$ $ \Longrightarrow$ $ P$ is on circle $ (U)$ with center $ U$ and radius $ UX = UZ.$ Similarly, $ P$ is on circle $ (V)$ with center $ V$ and radius $ VY = VT.$ $ XZ, YT$ are pairwise radical axes of $ (O), (U)$ and $ (O), (V)$ $ \Longrightarrow$ diagonal intersection $ W \equiv XZ \cap YT$ is radical center of $ (O), (U), (V)$ and $ PW$ is radical axis of $ (U), (V).$ Since $ (U), (V)$ are both perpendicular to $ (O),$ the circumcenter $ O \in PW.$ $ B \in YZ, D \in XT$ are pedals of $ P.$ Let $ B' \in YZ, D' \in XT$ be pedals of $ O$ and let $ B'' \in YZ, D'' \in XT$ be pedals of $ W.$ From collinearity of $ P, O, W$ and from $ \frac {YB'}{XD'} = \frac {YB''}{XD''} = \frac {YZ}{XT}$ $ \Longrightarrow$ $ \frac {YB}{XD} = \frac {YZ}{XT}.$ Let $ XY, ZT$ intersect at $ K$ and let $ (K), (P)$ be circles with centers $ K, P$ and equal radii $ KP = PK.$ Since polar of $ K$ goes through $ W,$ $ K$ is on polar $ UV$ of $ W$ $ \Longrightarrow$ circles $ (U), (V), (K)$ are coaxal and $ (K) \perp (O).$ Using sine theorem for $ \triangle UXK, \triangle VYK$ with $ \angle KXU = \angle KYV$ and $ \angle UKX = 180^\circ - \angle VKY,$ $ \frac {UP}{VP} = \frac {UX}{VY} = \frac {UK}{VK}$ $ \Longrightarrow$ $ PK$ bisects $ \angle UPV$ and circle $ (K)$ bisects angle formed by circles $ (U), (V).$ Inversion with center $ P$ and power $ PK^2$ takes circles $ (U), (V), (K)$ into concurrent lines $ u, v, k,$ such that $ k$ bisects $ \angle (u, v).$ Circle $ (O)$ perpendicular to $ (U), (V), (K)$ goes to a circle $ (O')$ centered at the concurrency point $ O'$ of $ u, v, k.$ Since $ (K) \cong (P),$ $ k$ is the perpendicular bisector of $ PK.$ Cyclic quadrilateral $ XYZT$ goes to rectangle $ X'Y'Z'T'$ with diagonal lines $ u \equiv X'Z', v \equiv Y'T'$ inscribed in $ (O')$ and with midline $ k \perp (X'Y' \parallel T'Z').$ By symmetry, $ \triangle T'PX' \cong \triangle Z'KY'$ are congruent $ \Longrightarrow$ $ \triangle XPT \sim \triangle Z'KY'$ are similar. On the other hand, $ \triangle ABC$ is pedal triangle of $ \triangle ZKY$ WRT $ P$ and $ \triangle Z'KY'$ is inversion image of $ \triangle ZKY$ WRT $ P$ $ \Longrightarrow$ $ \triangle ABC \sim \triangle Z'KY'$ are also similar (well known and posted before). As a result, $ \triangle XPT \sim \triangle ABC$ are similar. In exactly the same way, $ \triangle YPZ \sim \triangle ADC$ are also similar. Let $ \mathcal S_1: \triangle XPT \to \triangle ABC,$ $ \mathcal S_2: \triangle YPZ \to \triangle ADC$ be the corresponding similarity transformations. Due to $ \frac {XD}{XT} = \frac {YB}{YZ},$ these similarities map $ \mathcal S_1(D) \equiv \mathcal S_2(B)$ to the same point $ E \in AC.$ Since $ PD \perp XT, PB \perp YZ,$ it follows that $ BE \perp AC, DE \perp AC$ $ \Longrightarrow$ $ E \in BD$ is diagonal intersection of $ ABCD$ and $ BD \perp AC.$ $ \square$

Nice solution, Vladimir! I have a different proof, but I'm quite in a rush; so, I'll be a little "detail-less". We have the following results: Theorem 1. Let $ ABCD$ be a quadrilateral and let $ P = AC \cap BD$, $ E = AD \cap BC$, and $ F = AB \cap CD$. Then, $ \text{isog}_{ABE}{P} = \text{isog}_{CDE}{P} = \text{isog}_{ADF}{P} = \text{isog}_{BCF}{P}$ if and only if $ AC \perp BD$ (here $ \text{isog}_{XYZ}{P}$ means the isogonal conjugate of $ P$ with respect to triangle $ XYZ$). Theorem 2. Let $ ABCD$ be a quadrilateral and let $ P = AC \cap BD$. Denote by $ U1,\ V1,\ W1,\ T1$ the projections of $ P$ on $ AB,\ BC,\ CD,\ DA$ and by $ U2,\ V2,\ W2,\ T2$ the intersections of $ PU1,\ PV1,\ PW1,\ PT1$ with $ CD,\ DA,\ AB,\ BC,$ respectively. Then,$ ABCD$ is orthodiagonal if and only if $ U1,\ V1,\ W1,\ T1,\ U2,\ V2,\ W2,\ T2$ are all concyclic. The second is just a simple angle chase. Now by Theorem 2, and from the fact that two points have the same pedal circles wrt. a given triangle if and only if they are isogonally conjugated wrt. that triangle, we get Theorem 1. Now let $ Q = \text{isog}_{ABE}{P} = \text{isog}_{CDE}{P} = \text{isog}_{ADF}{P} = \text{isog}_{BCF}{P}$ and we shall see that this point satisfies the desired property.

Proof that $ P$ with desired properties exists $ \Longrightarrow$ $ AC \perp BD$ is actually even simpler than the reverse, just by an angle chase. Since $ ABCD$ is convex and the angles undirected, $ P$ must be inside of $ ABCD,$ otherwise some of the angles in the $ 90^\circ$ sums would be obtuse. Let $ F$ be intersection, other than $ P,$ of circumcircles $ \odot(PBC), \odot(PDA).$ Let $ E$ be intersection, other than $ F,$ of $ \odot(FAB), \odot(FCD).$ WLOG, let $ BFPC, AFPD$ follow on the corresponding circumcircles in this order. (Otherwise, relabel $ ABCD$ as $ DCBA.$) $ \angle AFB = 360^\circ - (\angle PFA + \angle PFB) = 360^\circ - (180^\circ - \angle PDA + 180^\circ - \angle PCB) =$ $ = \angle PDA + \angle PCB = 90^\circ.$ $ \angle CFD = \angle PFC + \angle PFD = \angle PCD + \angle PDA = 90^\circ.$ It follows that $ \angle AEB = \angle AFB = 90^\circ$ and $ \angle CED = \angle CFD = 90^\circ.$ The case $ P \equiv F$ (when $ BC \parallel DA$) does not present a problem, because then the angle chase can be repeated using the common tangent $ PF$ of the circles $ \odot(PBC), \odot(PDA).$ Assume that $ AEFB, DEFC$ follow on the corresponding circumcircles in this order. (Proof for the other order $ AFEB, DFEC$ is the same.) $ \angle DEA = 360^\circ - (\angle FED + \angle FEA) = 360^\circ - (180^\circ - \angle FCD + 180^\circ - \angle FBA) =$ $ = \angle FCP + \angle PCD + \angle PBA - \angle FBP = \angle PCD + \angle PBA = 90^\circ.$ $ \angle BEC = \angle FEB + \angle FEC = \angle FAB + \angle FDC =$ $ = \angle PAB - \angle PAF + \angle PDC + \angle PDF = \angle PAB + \angle PDC = 90^\circ.$ Again, the case $ E \equiv F$ (when $ AB \parallel CD$) does not present a problem, because then the angle chase can be repeated using the common tangent $ EF$ of the circles $ \odot(FAB), \odot(FCD).$ As a result, $ \angle AEB = \angle BEC = \angle CED = \angle DEA = 90^\circ.$ This means that $ E \equiv AC \cap BD$ is diagonal intersection and $ AC \perp BD.$ $ \square$

Suppose $ P$ exists and let $ M$ be the Miquel point for the complete quadrilateral formed by the lines $ AB, BC, CD, DA$. Perform an inversion of center $ M$. Let $ Q = AB \cap CD, R = BC \cap DA$. The circumcircles of $ \triangle QAB$ and $ \triangle PAB$ are orthogonal by the problem condition, thus the image of the circle $ (PAB)$ is a circle of diameter $ A'B'$. Analogous reasoning holds for $ (PBC), (PCA)$, etc. therefore $ P'$ is the intersection of the perpendicular $ A'C'$ and $ B'D'$. But $ \angle AMB = \angle CMB \Rightarrow$ the angle bisectors of $ \angle AMC$ and $ \angle BMD$ coincide $ \Rightarrow \angle (AC, BD) = \angle (B'D', A'C')$. And conversely.

Existence of $P$ $\implies AC\perp BD$: We show that $AB^2+CD^2=BC^2+AD^2$, which implies that $AC\perp BD$. Let $PA = a, PB = b, PC = c, PD = d$, and $\angle PDA = \theta_1, \angle PAD = \theta_2$, so $\angle PCB = \frac{\pi}{2}-\theta_1, \angle PBC = \frac{\pi}{2}-\theta_2$. Note that all the mentioned angles above are acute ($P$ is inside the quadrilateral). More importantly, the $\sin$ and $\cos$ of all these angles are positive. Note that $\angle APD+\angle BPC = \pi, \angle APB + \angle CPD = \pi$. So by the Law of Cosines, \[AD^2+BC^2=a^2+b^2+c^2+d^2-2(ad-bc)\cos{\angle APD}\] and \[AB^2+CD^2=a^2+b^2+c^2+d^2-2(ab-cd)\cos{\angle APB}\]. So $AB^2+CD^2=AD^2+BC^2 \Longleftrightarrow (ad-bc)\cos{\angle APD} = (ab-cd)\cos{\angle APB}$. We proceed to compute $\cos{\angle APD}$. By the Law of Sines, \[\frac{a}{d}\cdot \sin{\theta_2} = \sin{\theta_1}\]\[\frac{b}{c}\cdot {\cos{\theta_2}}=\cos{\theta_1}\]. Squaring the 2 above equations, and summing, we get that \[\frac{a^2}{d^2}\cdot\sin^2{\theta_2}+\frac{b^2}{c^2}\cdot \left(1-\sin^2{\theta_2}\right) = 1 \implies \sin{\theta_2} = d \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \implies \cos{\theta_2} = c \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\] (the sine and cosine are both positive). Now we easily get that \[\sin{\theta_1} = a \cdot \sqrt{\frac{c^2-b^2}{a^2c^2-b^2d^2}} \hspace{1cm} \cos{\theta_1} = b \cdot \sqrt{\frac{a^2-d^2}{a^2c^2-b^2d^2}}\]. Now note that \[\cos{\angle APD} = -\cos({\theta_1+\theta_2}) = -(\cos{\theta_1}\cos{\theta_2}-\sin{\theta_1}\sin{\theta_2}) \\ = -\frac{a^2bc-bcd^2}{a^2c^2-b^2d^2}+\frac{ac^2d-ab^2d}{a^2c^2-b^2d^2} \\= -\frac{(ac+bd)(ab-cd)}{(ac+bd)(ac-bd)}\\ = -\frac{ab-cd}{ac-bd} \\ \implies (ad-bc)\cos{\angle APD} = -\frac{(ab-cd)(ad-bc)}{ac-bd}\]. Since this is symmetric, $(ad-bc)\cos{\angle APD} = (ab-cd)\cos{\angle APB}$. Note that I still have dumb cases to deal with, like $a = d, b = c$, but those are silly. $AC \perp BD \implies$ existence of $P$: Let $AC \cap BD = O$. We show that the isogonals of $AO, BO, CO, DO$, with respect to $\angle BAD, \angle ABC, \angle BCD, \angle CDA$, concur at our desired point $P$. By simple angle chasing, it is obvious that this $P$ satisfies the properties. Let the isogonals of $AO, BO, CO, DO$ be $%Error. "fancy" is a bad command. {l}_a, %Error. "fancy" is a bad command. {l}_b, %Error. "fancy" is a bad command. {l}_c, %Error. "fancy" is a bad command. {l}_d$. Let $\angle({line_1, line_2})$ denote the angle between 2 lines (positive or negative). Finally, let $\angle BAC, \angle CAD, \angle ACB, \angle ACD$ be $\theta_1, \theta_2, \theta_3, \theta_4$ respectively. We show that $%Error. "fancy" is a bad command. {l}_a, %Error. "fancy" is a bad command. {l}_b, %Error. "fancy" is a bad command. {l}_c$ concur. The other triplet follows similarly. We use Ceva's theorem (sine form) for this. Note that $\angle(AC, %Error. "fancy" is a bad command. {l}_a) = \theta_1-\theta_2, \angle(%Error. "fancy" is a bad command. {l}_a, AB) = \theta_2, \angle(AB, %Error. "fancy" is a bad command. {l}_b) = \frac{\pi}{2} - \theta_3, \angle(%Error. "fancy" is a bad command. {l}_b, BC) = \frac{\pi}{2}-\theta_1, \angle(BC, %Error. "fancy" is a bad command. {l}_c) = \theta_4, \angle(%Error. "fancy" is a bad command. {l}_c, AC) = \theta_3-\theta_4$. So \[\frac{\sin(\theta_1-\theta_2)}{\sin{\theta_2}}\cdot\frac{\sin(\frac{\pi}{2} - \theta_3)}{\sin(\frac{\pi}{2}-\theta_1)}\cdot{\frac{\sin{\theta_4}}{\sin(\theta_3-\theta_4)} = \frac{\sin{\theta_1}\cos{\theta_2}-\sin{\theta_2}\cos{\theta_1}}{\sin{\theta_2}}\cdot\frac{\cos{\theta_3}}{\cos{\theta_1}}\cdot{\frac{\sin{\theta_4}}{\sin{\theta_3}\cos{\theta_4}-\sin{\theta_4}\cos{\theta_3}}}}\] \[\\ = \frac{\frac{OB}{AB}\cdot\frac{OA}{AD}-\frac{OD}{AD}\cdot\frac{OA}{AB}}{\frac{OD}{AD}}\cdot\frac{\frac{OC}{BC}}{\frac{OA}{AB}}\cdot{\frac{\frac{OD}{DC}}{\frac{OB}{BC}\cdot\frac{OC}{CD}-\frac{OD}{DC}\cdot\frac{OC}{BC}} = 1}\] (after elementary calculation). So the result follows.

If $AC\perp BD$ lets consider $P$ to be the isogonal conjugate point of $E=AC\cap BD$ wrt $RAB$ ($R=BC\cap AD$). Now let's consider the orthocenter $X$ of $CDA$. We have $BCX\sim BPA$($\angle CXB=\angle CAD=\angle BAP$ and $\angle PBA=\angle XBC$) so $BX\cdot BP=BA\cdot BC$ Now from this we have $BAX\sim BPC$ giving $\angle DCA=\angle BXA=\angle BCP$ similarly we have that $\angle PDA=\angle BDC$ and it's easy to check that $P$ satisfies all the conditions. Now assume such $P$ exists. if again $R=AD\cap BC$ we consider $K,M$ which are isogonal conjugates of $P$ wrt $RAB$ and $RCD$. The condition for angles gives that lines $AK,KB,CM,DM$ form a rectangle now note that if $AK,BK$ cut $BC,AD$ at $C',D'$ we have that $RCD$ and $RC'D'$ are homothethic(because $\angle KRA=\angle PRB=\angle MRA$ meaning that $K,M,R$ are collinear). But using the if part(on $ABC'D'$ we know that $P$ is the isogonal conjugate of $K$ wrt triangle $AC'D'$ but now since $P$ is the isogonal conjugate of $M$ wrt $ACD$ we have that the homothethy that sends $ACD$ to $AC'D'$ (and $M$ to $K$) sends $P$ to $P$ this gives $M=K$ which means $AC\perp BD$

Well known: In triangle ABC if P and Q are isogonal conjugates, the midpoint of PQ is the center of the circle that passes through the feet of the perpendiculars from P and Q to the sides. Let AB and CD cut at F, and AD and BC cut at E. Assume the diagonals are perpendicular at point X. From X draw the perpendiculars to the sides, and these 4 points are concyclic (easy to see by anglechase). Define P to be the reflection of X across the center of this circle. Then by the well-known theorem, P and X are isogonal conjugates in triangles ABE, DCE, FAD, FBC and the angle condition is immediately obtained. Now assume the angle condition is true, then do exactly the same but switching P and X.

Nice problem! Part 1. :- Let $AC \perp BD$ Proof Indeed, let $AC,BD$ meet at point $R$, lines $AB,CD$ and $AD,BC$ meet at points $Q,P$ respectively. Let the reflections of $P$ in $AB,BC,CD,DA$ be $X,Y,Z,T$ respectively. By some old USAMO problem, whose link I shall post later on, we know that $X,Y,Z,T$ are con cyclic. Let $S$ be the centre of this circle. It is evident that $S$ is the isogonal conjugate of $R$ in triangles $PAB,PDC,QAD,QBC$. Thus, we easily see that our result is satisfied by point $S$ Part 2. :- Let such a point $X$ exist Proof Indeed, doing as previously, we let the reflections of our point $S$ be $X,Y,Z,T$. Then, by our angle condition, these points are infact con cyclic. Thus, if we let $R'$ be the centre, then it is the isogonal conjugate of $S$ in all these four triangles. Therefore, we see that $AB,BC,CD,DA$ sub tend the same right angle at $R'$ and so $R'=R$ which means that $AC \perp BD$.

Here's an overkill solution:

Solution from Twitch Solves ISL: The problem is killed by quoting the following theorem: Theorem: If $X$ is a point such that $\angle AXD + \angle BXC = 180^{\circ}$, then the isogonal conjugate of $X$ exists. Proof. [Proof, for completeness only] It's enough for the projections of $X$ to the sides to be cyclic, by considering the six-point circle. Let them be $X_1$, $X_2$, $X_3$, $X_4$. \begin{align*} \measuredangle XX_4X_1 &= \measuredangle XAX_1 \\ &\vdots \end{align*}and this means $\measuredangle X_1X_2X_3 = \measuredangle X_1X_4X_3$. $\blacksquare$ Now in one direction, if the diagonals are perpendicular and meet at $Q$, then its isogonal conjugate exists, and is seen to have the desired property. Conversely, given such a point $P$, it has an isogonal conjugate $Q$ which satisfies $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90^{\circ}$, which implies that $Q$ is the perpendicular intersection of the diagonals.

lol wut. By the angle condition $\angle APD+\angle CPB=180$, so $P$ has an isogonal conjugate $P'$. However we then have $$\angle AP'D=\angle DP'C=\angle CP'B=\angle BP'A=90$$ So the diagonals are perpendicular. The reverse condition is true by taking the isogonal conjugate of $AC\cap BD$. $\square$

If $AC\perp BD$ point $E=AC\cap BD$ has isogonal conjugate $Q$ in $ABCD,$ and clearly $P=Q$ works. If $P$ exists, then $\angle APB+\angle CPD=\pi,$ so $P$ has isogonal conjugate $F$ in $ABCD$ and $$\angle AFB=\angle BFC=\angle CFD=\angle DFA=\frac{\pi}{2}\implies F=E\implies AC\perp BD.$$

Let $E=AC\cap BD$, since $\angle AEB+\angle CED=180^{\circ}$ there exists an isogonal conjugate $E'$ which obviously satisfies the conditions. (I guess if you wanted to prove the isogonal conjugate lemma you could draw the (cyclic) pedal quadrilaterals of $E$ and $E'$ and then do some PoP, etc.)

Perhaps I should be a bit more specific when proving constructions using geometric continuity.

, it is easy to see that there exists a rectangle with two sides parallel to each diagonal of $ABCD$ such that it is a pedal quadrilateral of $ABCD$. Thus we're done.

Observe that $P$ has a pedal circle. We will prove a known lemma: $\textbf{Claim.}$ If $P$ has a pedal circle in a convex $n$ gon then it also has an isogonal conjugate. $\textit{Proof.}$ Reflect $P$ over the circumcenter of its pedal circle, and then this point is an isogonal conjugate of $P$ in all the triangles formed by $3$ adjacent sides. $\square$ Take the isogonal conjugate of $P$, call it $Q$, and the conditions rewrite to being $\angle AQB = \angle BQC = \angle CQD = \angle DQA = 90^\circ.$ Thus $B-Q-D$ and $A-Q-C$, and hence $AC \perp BD$. For the converse, it is known that the intersection of the diagonals has a pedal circle. Thus take the isogonal conjugate to finish.

Let $Q = AC \cap BD$. For the forwards direction, note $Q$ has an isogonal conjugate, and this point satisfies the conditions of our desired $P$. For the backwards direction, note $P$ has an isogonal conjugate, which can be determined to be $Q$. $\blacksquare$

We show the following useful lemma first. Lemma. In quadrilateral $ABCD$ a point $P$ has a pedal circle iff it has an isogonal conjugate. Proof. If $P$ has a pedal circle then the reflection $Q$ of $P$ over its center will be its isogonal conjugate with respect to the triangle with sides lines $AB,BC,CD$ by a property of isogonal conjugation in triangles, and thus $PB,QB$ and $PC,QC$ are isogonal in both this triangle and $ABCD.$ Similarly $Q$ is the isogonal conjugate of $P$ in the triangle with sides lines $AB,DA,CD$ and thus $PA,QA$ and $PD,QD$ are isogonal in $ABCD.$ If $P$ has an isogonal conjugate $Q$ then $P,Q$ are isogonal conjugates in the triangle with sides lines $AB,BC,CD$ and thus the midpoint of $PQ$ is equidistant to the feet from $P$ to $AB,BC,CD.$ Similarly it is equidistant to the feet from $P$ to $AB,DA,CD$ so $P$ has a pedal circle centered at the midpoint of $PQ.$ Now let $ABCD$ be orthodiagonal and let $Q$ be the intersection of its diagonals. By 1993 USAMO 2, $Q$ has a pedal circle and thus an isogonal conjugate $P$ which we can see will satisfy the desired angle conditions. Now let $P$ be a point in any $ABCD$ satisfying the conditions. Let $E,F,G,H$ be the feet from $P$ to $DA,AB,BC,CD$ respectively. We see \[\angle EFG=\angle EFP+\angle PFG=\angle EAP+\angle PBG=\angle PBC+\angle PAD=90^\circ,\]and similarly $\angle FGH=\angle GHE=\angle HEF=90^\circ.$ In particular $EFGH$ is cyclic. Thus by our lemma $P$ has some isogonal conjugate $Q.$ We can check that the angle conditions imply $\angle AQB=\angle BQC=\angle CQD=\angle DQA=90^\circ,$ so $Q$ is the intersection of $AC,BD$ and thus they are perpendicular.

We claim that if $X$ is a point inside quadrilateral $ABCD$ such that $\angle AXB+\angle CXD=180^\circ$ then $X$ has an isogonal conjugate. Let the feet of the altitudes from $X$ to $AB$, $BC$, $CD$, $DA$ be $X_1$, $X_2$, $X_3$, $X_4$ respectively. Note that $AX_1XX_4$, $BX_2XX_1$, $CX_3XX_2$, and $DX_4XX_3$ are concyclic. Then, we have \begin{align*} \angle X_1X_4X_3 &= 180^\circ-\angle AX_4X_1-\angle DX_4X_3\\ &= 180^\circ - \angle AXX_1-\angle DXX_3 \angle X_1X_2X_3&=180^\circ-\angle X_1X_2B-\angle X_3X_2C\\ &= 180^\circ - \angle BXX_1-\angle CXX_3 \end{align*}but since $\angle AXX_1+\angle BXX_1+\angle CXX_3+\angle DXX_3=180^\circ$, we have that $X_1X_2X_3X_4$ is also cyclic. Now, we claim that this means $X$ has an isogonal conjugate. Now, reflect $X$ over the center of the circle to $Y$. We claim that $Y$ is the isogonal conjugate. Note that when you drop the same perpendiculars from $Y$, we can work backwards to get the same conclusions. The rest of the problem is trivial. If the conditions are true then $P$ has an isogonal conjugate and is where the diagonals are perpendicular, and if the diagonals are perpendicular, their intersection point has an isogonal conjugate that is $P$.

Let $AC \cap BD = Q$. We start by proving the backwards direction. Note that $Q$ has an isogonal conjugate since $\angle AQB + \angle CQD = 180^\circ$. Then let $P'$ be the isogonal conjugate of $Q$. Then it follows that since $\angle QAB + \angle QBA = 90^\circ$, we have $\angle PAD + \angle P'BC = 90^\circ$, and repeating gives us the problem condition so $P' = P$ as desired. The forward condition is the same as we can directly take the isogonal conjugate of $P$ which clearly exists, so we are done.