Very nice. Note that the pedal circle of a point $X$ is also the pedal circle of its isogonal conjugate. Let $R'=\overline{PQ'}\cap\overline{P'Q}$. Then $R,R'$ are isogonal conjugates by DDIT from $A,B,C$ to complete quadrilateral $\{\overline{PQ},\overline{P'Q},\overline{PQ'},\overline{P'Q'}\}$. Let the projections of these onto $\overline{BC}$ be $P_A,P'_A,Q_A,Q'_A,R_A,R'_A$. Let the point at infinity perpendicular to $\overline{BC}$ be $P_\infty$. Projecting the involution created by DDIT with $P_{\infty}$ and $\{\overline{PQ},\overline{P'Q},\overline{PQ'},\overline{P'Q'}\}$ onto $\overline{BC}$ means that $(P_A,P'_A),(Q_A,Q'_A),(R_A,R'_A)$ are pairs under an involution, so there is some point $Y_A$ on $\overline{BC}$ which has equal power to all $3$ circles. Similarly there are points $Y_B$ and $Y_C$ with this property, and not all 3 of these can be the same, so there are $2$ distinct points with equal power to all $3$ circles, as desired.

Ratio of the power of the projection of $R$ onto $BC$ to $(P)$ and $(Q)$ is $\frac{RP' \cdot \cos (P'Q', BC) \cdot RP \cdot \cos (PQ, BC)}{RQ' \cdot \cos (P'Q', BC) \cdot RQ \cdot \cos (PQ, BC)} = \frac{RP \cdot RP'}{RQ \cdot RQ'}$. Thus all projections of $R$ onto the sides have the same power ratio. We are done by power ratio lemma.

Let $R'$ be the intersection of $PQ'$ with $P'Q$. Applying the Isogonal Lines Lemma to $\angle ABC$ with conjugates $(P,P')$ and $(Q,Q')$ we get that $BR$ and $BR'$ are isogonal lines in $\angle ABC$. Applying cyclically gives that $R$ and $R'$ are isogonal conjugates. It is well known that the center of the pedal circles are the midpoints of $PP'$, $QQ'$, and $RR'$ which all lie on the Newton-Gauss line of complete quadrilateral $PP'QQ'RR'$.

sami1618 wrote: Let $R'$ be the intersection of $PQ'$ with $P'Q$. Applying the Isogonal Lines Lemma to $\angle ABC$ with conjugates $(P,P')$ and $(Q,Q')$ we get that $BR$ and $BR'$ are isogonal lines in $\angle ABC$. Applying cyclically gives that $R$ and $R'$ are isogonal conjugates. It is well known that the center of the pedal circles are the midpoints of $PP'$, $QQ'$, and $RR'$ which all lie on the Newton-Gauss line of complete quadrilateral $PP'QQ'RR'$. Proving that their centers are collinear is not enough to conclude that they are coaxal. You need to find the existence of at least one ordinary point (not at infinity) that has equal power to them. i3435, in post #1, proved the existence of 3, which already yields the desired conclusion leaving no respect for that Newton line. Just for the record, the orthopole of $PQ$ has equal power WRT them, therefore it is on their common radical axis. See radical axis and Simpson's lines are concurrent.

Generalization : Given a $ \triangle ABC $ with orthocenter $ H $ and two pairs of isogonal conjugate $ (P, P^{*}), (Q, Q^{*}). $ Let $ \Omega_{P}, \Omega_{Q} $ be the pedal circle of $ (P, P^{*}), (Q, Q^{*}) $ WRT $ \triangle ABC, $ respectively. Then the radical axis of $ \Omega_{P}, \Omega_{Q} $ is the image of the Steiner line of $ PQP^{*}Q^{*} $ under the dilation $ (H, \tfrac{1}{2}). $ Proof : Let $ S $ be a point on the Steiner line of $ PQP^{*}Q^{*} $ and $ M, T $ be the midpoint of $ PP^{*}, HS, $ respectively, then$\color{blue} ^{[1]}$ \begin{align*} 4\cdot \overline{TM}^2 &= 2\cdot\overline{TP}^2\ +\ 2\cdot\overline{TP^{*}}^2\ -\ \overline{PP^{*}}^2\\ &= \overline{HP}^2\ +\ \overline{HP^{*}}^2\ +\ \overline{SP}^2\ +\ \overline{SP^{*}}^2\ -\ \overline{PP^{*}}^2\ -\ \overline{HS}^2\\ &= \mathbf{Power}(H, \odot (O))\ +\ 4 \cdot {r_{P}}^{2}\ -\ 2 \cdot \mathbf{Power}(S, \odot (PP^{*}))\ -\ \overline{HS}^2, \end{align*}where $ r_{P} $ is the radius of $ \Omega_{P}, $ so note that $ \mathbf{Power}(S, \odot (PP^{*})) = \mathbf{Power}(S, \odot (QQ^{*})) $ we conclude that \begin{align*} 4\cdot\mathbf{Power}(T, \Omega_{P}) &= 2 \cdot \mathbf{Power}(S, \odot (PP^{*})) + \overline{HS}^2 - \mathbf{Power}(H, \odot (O))\\ &= 2 \cdot \mathbf{Power}(S, \odot (QQ^{*})) + \overline{HS}^2 - \mathbf{Power}(H, \odot (O))\\ &=4\cdot \mathbf{Power}(T, \Omega_{Q}). \qquad \blacksquare \end{align*}[1] Blog-A cute problem about Radical Axis (Lemma)