Let $H_1,H_2,H_3,H_4$ be the orthocenters of $\triangle APD,$ $\triangle AQB,$ $\triangle CQD,$ $\triangle BPC.$ $H_1,H_2,H_3,H_4$ are collinear on the Steiner line of the complete quadrangle $ABCD,$ thus it suffices to prove that $\overline{H_1H_4}$ and $\overline{H_2H_3}$ have the same midpoint. Since $AB,CD$ are antiparallel with respect to $QD,QC,$ then $QH_2 \perp AB$ is the Q-circumdiameter of $\triangle QDC,$ which cuts its circumcircle again at the reflection $F$ of its orthocenter $H_3$ about the midpoint $M$ of $CD.$ Hence, if $T$ is the midpoint of $\overline{H_1H_2},$ then $MT \parallel FH_2,$ i.e. $MT \perp AB$ $\Longrightarrow$ CD-maltitude of $ABCD$ passes through the midpoint $T$ of $\overline{H_2H_3}.$ Similarly, AB-maltitude of $ABCD$ passes through $T$ $\Longrightarrow$ $T$ is the anticenter of $ABCD.$ Likewise, $T$ is the midpoint of $\overline{H_1H_4}$ $\Longrightarrow$ $H_1H_2=H_3H_4.$