I will just sketch the proof of this beautiful problem. Let $A',B',C',E',F',P'$ be the midpoints of $BC,AC,AB,PC,BP,AP$ respectively. 1- $B'F'$, $C'E'$ and $P'A'$ meet at a single point $G$. Morevoer $G$ is the midpoint of these three segments. This is easy to prove, just look for certain parallelograms. 2- Let $H$ be the reflection of $O$ through $G$. $OB'EF'$ is cyclic and $O$ is the antípode of $E$, then $H$ is the orthocenter of $\triangle B'EF'$. Analogously $H$ is the orthocenter of $\triangle C'E'F$ and $\triangle P'A'D$. 3- Now it is easy to realize that $Q$ is just the midpoint of $GH$. 4- Now the fun part, when $P$ moves along $(ABC)$ homothecy from $A$ shows that $P'$ describes a circle, then homothecy from $A'$ shows that $G$ describes a circle, then homothecy from $O$ shows that $Q$ describes a circle.

This proof is quite the same as the previous one, but let it be here for the storage. Proof. Let $H$, $G$ be the orthocenter and centroid of $\triangle ABC$, respectively. Let $A_1$, $A_2$ be the midpoints of $\overline{AP}$, $\overline{BC}$, respectively, define other points similarly. It's well-known that the midpoints of $\overline{A_1A_2}$, $\overline{B_1B_2}$, $\overline{C_1C_2}$ coincide (this holds for any quadruple of points $A,B,C,P$ in the plane, which is easily proven by midlines and parallelograms). Let this common midpoint be $T$. By Menelaus's for $\triangle AA_1A_2$ and points $G,T,P$ on its sides we get that $G,T,P$ are collinear. Then by Menelaus's theorem for $\triangle GPA$ and points $A_1,T,A_2$ on its sides we get that $\frac{PT}{GT}=3$. Let $S$ be the midpoint of $\overline{PH}$ and let $O$ be the reflection of $S$ over $T$. By Menelaus's theorem for $\triangle HOS$ and points $G,T,P$ on its sides we get that $\frac{HG}{GO}=2 \implies O$ is the circumcenter of $\triangle ABC$. Since obviously $O$ is the antipode of $D$ on $(DA_1A_2)$, it's well-known that $S$ is the orthocenter of $\triangle DA_1A_2 \implies S \in \overline{DX}$ and $\angle SXT=90^{\circ}$. From the symmetry $(XYZ)$ is the circle with diameter $\overline{ST}$, and $Q$ is its midpoint. Since $(ABC)$ gets sent to the nine-point circle of $\triangle ABC$ under homothety at $H$ with ratio $\frac{1}{2}$, $S \in \text{NPC}$. Let $N$ be the midpoint of $\overline{OH}$, which is the center of $\text{NPC}$, and $L$ be the midpoint of $\overline{ON}$. Then if we consider such positions of $P$ which lie both on $(ABC)$ and the Euler line of $\triangle ABC$, using some length bashing from the written above we get that $LT$ is equal to $\frac{1}{2}$ of the radius of the $\text{NPC}$, which is $\frac{1}{4}$ of the radius of $(ABC)$. If we consider homothety at $G$ with ratio $\frac{1}{4}$, then $P$ gets sent to $T$, thus as $P$ moves along $(ABC)$, $T$ moves along the circle with center $L$ and radius $\frac{1}{4}$ of that of $(ABC)$. Proceeding in the same manner as we did in the previous paragraph, we prove that $Q$ moves along the circle with center being the midpoint of $\overline{LN}$ and radius equal to $\frac{3}{8}$ of that of $(ABC)$, and we're finally done. $\blacksquare$