Assuming $ AB \parallel CD$ $ \Longrightarrow$ isosceles $ \triangle ABP \sim \triangle CDP$ are centrally similar with center $ P$ $ \Longrightarrow$ $ ABCD$ is isosceles trapezoid with diagonal intersection $ P,$ with perpendicular diagonals $ AC \perp BD,$ cyclic, and $ Q$ is obviously its circumcenter. Assuming $ BC \parallel DA$ $ \Longrightarrow$ isosceles $ \triangle ABP \cong \triangle CDP$ are congruent $ \Longrightarrow$ $ ABCD$ is isosceles trapezoid with circumcenter $ P,$ with perpendicular diagonals $ AC \perp BD$ and $ Q$ is obviously its diagonal intersection. Hence, we can assume $ AB, CD$ are not parallel and $ BC, DA$ are also not parallel. Let $ K, M$ be midpoints of $ AB, CD.$ Let $ X \equiv AB \cap CD,$ WLOG on the same side of $ KM$ as $ B, C.$ Quadrilateral $ PKXM$ is cyclic on account of the right angles at $ K, M$; let $ (S)$ be its circumcircle. Let $ PB, PC$ cut $ (S)$ again at $ B', C'.$ By Pascal theorem for cyclic hexagon $ PKB'XC'M,$ the intersections $ B \equiv PB' \cap KX,$ $ C \equiv PC' \cap MX,$ $ L \equiv KC' \cap MB'$ are collinear $ \Longrightarrow$ $ L \in BC.$ Then $ \angle KML = \angle KMB' = \angle KPB' = \angle KPB = 45^\circ$ $ \angle MKL = \angle MKC' = \angle MPC' = \angle MPC = 45^\circ$ Let $ PA, PD$ cut $ (S)$ again at $ A', D'.$ By Pascal theorem for cyclic hexagon $ PKD'XA'M,$ the intersections $ A \equiv PA' \cap KX,$ $ D \equiv PD' \cap MX,$ $ N \equiv KD' \cap MA'$ are also collinear $ \Longrightarrow$ $ N \in DA.$ Then $ \angle KMN = 180^\circ - \angle KMA' = 180^\circ - \angle KPA' = \angle KPA = 45^\circ$ $ \angle MKN = 180^\circ - \angle MKD' = 180^\circ - \angle MPD' = \angle MPD = 45^\circ$ Combining, $ \triangle KLM \cong \triangle KNM$ are isosceles right and congruent $ \Longrightarrow$ $ KLMN$ is a square. Assume that $ L' \not\equiv L, N'\not\equiv N$ are midpoints of $ BC, DA.$ Since $ KLMN, KL'M'N'$ are both parallelograms $ \Longrightarrow$ $ KL = MN, KL' = MN', \angle LKL' = \angle NMN'$ $ \Longrightarrow$ $ \triangle LKL' \cong \triangle NMN'$ are centrally congruent by s.a.s. and $ (BC \equiv LL') \parallel (NN' \equiv DA),$ which is a contradiction. Therefore, $ L, N$ are midpoints of $ BC, DA$ and the square $ KLMN$ is Varignon parallelogram of the quadrilateral $ ABCD.$ Consequently, $ ABCD$ has perpendicular diagonals $ (AC \parallel KL) \perp (LM \parallel BD).$ The diagonal intersection $ E \equiv AC \cap BD$ is the 2nd intersection of the circles $ (K) \equiv \odot (ABP), (M) \equiv \odot (CDP).$ Let $ Q$ be the 2nd intersection of the circles $ (L) \equiv \odot(BCE), (N) \equiv \odot (DAE)$ $ \Longrightarrow$ the angles $ \angle BQC, \angle DQA$ are right. Radical axis $ EQ$ of $ (L), (N)$ is perpendicular to their center line, $ EQ \perp LN.$ Moreover, $ (LN \parallel BD) \perp AC$ or $ LN \perp (PC \equiv PA)$ $ \Longrightarrow$ $ \angle QBC = \angle QEC = \angle MLN = 45^\circ$ $ \Longrightarrow$ $ \triangle BCQ$ is isosceles right. $ \angle QAD = \angle QED = \angle MNL = 45^\circ$ $ \Longrightarrow$ $ \triangle ADQ$ is isosceles right.

Here is shorter solution to the problem: Let's denote by $ O$ the point on $ BD$ such that $ AO$ is perpendicular to $ BD$. Than, $ \angle{AO_{1}B} = 90^\circ = \angle{APB}$. It means that $ A,P,O_{1},B$ are concyclic. So, $ \angle{AO_{1}P} = \angle{APB} = 45^\circ$ and $ = \angle{DO_{1}P} = \angle{DCP} = 45^\circ$ and points $ D,C,O_{1},P$ are also concyclic, yielding $ \angle{DO_{1}C} = \angle{DPC} = 90^\circ$ and $ \angle{AO_{1}B} + \angle{CO_{1}D} = 180^\circ$ and hence, $ O$ is the intersection of $ AC$ and $ BD$ and the diagonals are perpendicular to each other. Furthermore, $ \triangle{APC} = \triangle{BPD}$ and $ AC = BD$ (*). Let $ Q$ be the common point of circumcircles of triangles $ AOB$ and $ COD$, other than $ O$. $ \angle{ADQ} = \angle{COQ} = \angle{CBQ}$. Hence, triangles $ ADQ$ and $ CBQ$ are similar and $ \frac {BQ}{CQ} = \frac {QD}{QA}$, but $ \angle{BQD} = \angle{CQA}$. So, $ \triangle{BQD}$ is similar to $ \triangle{CQA}$, but (*) yields that they are equal and $ BQ = QC$, $ AQ = QD$. So there exists such a point $ Q$. Remark: If the circumcircles of triangles $ AOB$ and $ COD$ are tangent to each other, than we analogously deduce that $ O = Q$ satisfies the problem.

Short complex numbers solution: Let P be the origin. A = a, B = ai, C = c, D = ci. Let Z=z the the point such that (b-z)i = c-z, or equivalently a+c-z=-zi, which can be written as (ci-z)i = (d-z)i = a-z. Thus, z = Q.

SKIBIDIRIZZTASTIC!!!! Note that if P exists and has an isogonal conjugate, then Q will exist and satisfy the conditions. This only happens if the pedal circle of P exists: now clearly this is true from an easy angle chase

It is well known that a point $P$ has an isogonal conjugate in quadrilateral $ABCD$ if $\angle APB + \angle CPD=180$ which is satisfied here. Hence $Q$ is the required point when it is the isogonal conjugate of $P$.