Solution:

We write $\omega_{1}, \omega_{2}$ and $\omega'$ for the circumcircles of $AGD, AEF$ and $OO_{1}O_{2}$ respectively. Since $O_{1}$ and $O_{2}$ are the centers of $\omega_{1}$ and $\omega_{2}$, and because $DG$ and $EF$ are parallel, we get that
$$\angle GAO_{1} = 90^{\circ} - \frac{\angle GO_{1}A}{2} = 90^{\circ} - \angle GDA = 90^{\circ} - \angle EFA = 90^{\circ} - \frac{\angle EO_{2}A}{2} = \angle EAO_{2}$$
So, because $G, A$ and $E$ are collinear, we come to the conclusion that $O_{1}, A$ and $O_{2}$ are also collinear.

Let $\angle DFE = \varphi$. Then, as a central angle $\angle AO_{2}E = 2\varphi$. Because $AE$ is a common chord of both $\omega$ and $\omega_{2}$, the line $OO_{2}$ that passes through their centers bisects $\angle AO_{2}E$, thus $\angle AO_{2}O = \varphi$. By the collinearity of $O_{1}, A, O_{2}$, we get that $\angle O_{1}O_{2}O = \angle AO_{2}O = \varphi$. As a central angle in $\omega'$, we have $\angle O_{1}PO = 2\varphi$, so $\angle POP_{1} = 90^{\circ} - \varphi$. Let $Q$ be the point of intersection of $DF$ and $OP$. Because $AD$ is a common chord of $\omega$ and $\omega_{1}$, we have that $OO_{1}$ is perpendicular to $DA$ and so $\angle DQP = 90^{\circ} - \angle POP_{1} = \varphi$. Thus, $OP$ is parallel to $\ell_{C}$ and so to $\ell_{B}$ as well.

Alternative Solution by PSC.

Let us write $\alpha, \beta, \gamma$ for the angles of $ABC$. Since $ADBC$ is cyclic, we have $\angle GDA = 180^{\circ} - \angle BDA = \gamma$. Similarly, we have
$$\angle GAD = 180^{\circ} - \angle DAE = \angle EBD = \angle BEC = \angle BAC = \alpha$$
where we have also used the fact that $\ell_{B}$ and $\ell_{C}$ are parallel.
Thus, the triangles $ABC$ and $AGD$ are similar. Analogously, $AEF$ is also similar to them.

Since $AD$ is a common chord of $\omega$ and $\omega_{1}$ then $AD$ is perpendicular to $OO_{1}$. Thus,
$$\angle OO_{1}A = \frac{1}{2} \angle DO_{1}A = \angle DGA = \beta$$
Similarly, we have $\angle OO_{2}A = \gamma$. Since $O_{1}, A, O_{2}$ are collinear (as in the first solution) we get that $OO_{1}O_{2}$ is also similar to $ABC$. Their circumcentres are $P$ and $O$ respectively, thus $\angle POO_{1} = \angle OAB = 90^{\circ} - \gamma$.
Since $OO_{1}$ is perpendicular to $AD$, letting $X$ be the point of intersection of $OO_{1}$ with $GD$, we get that $\angle DXO_{1} = 90^{\circ} - \gamma$. Thus $OP$ is parallel to $\ell_{B}$ and therefore to $\ell_{C}$ as well.

Solution:

Let $L$ and $Z$ be the points of intersection of $OO_{1}$ with $\ell_{b}$ and $DA$ respectively. Since $LZ$ is perpendicular on $DA$, and since $\ell_{b}$ is parallel to $\ell_{c}$, then
$$\angle DLO = 90^{\circ} - \angle LDZ = 90^{\circ} - \angle DFE = 90^{\circ} - \angle AFE$$
Since $AE$ is a common chord of $\omega$ and $\omega_{2}$, then it is perpendicular to $OO_{2}$. So letting $H$ be their point of intersection, we get
$$\angle DLO = 90^{\circ} - \angle AFE = 90^{\circ} - \angle AO_{2}H = \angle O_{2}AH$$
Let $K, Y, U$ be the projections of $P$ onto $OO_{2}, O_{1}O_{2}$ and $OO_{1}$ respectively. Then $YKUO_{1}$ is a parallelogram and so the extensions of $PY$ and $PU$ meet the segments $UK$ and $KY$ at points $X, V$ such that $YX \perp KU$ and $UV \perp KY$.
Since the points $O_{1}, A, O_{2}$ are collinear, we have
$$\angle FAO_{2} = O_{1}AZ = 90^{\circ} - \angle AO_{1}Z = 90^{\circ} - \angle YKU = \angle PUK = \angle POK = \angle POK$$
where the last equality follows since $PUOK$ is cyclic.
Since $AZOH$ is also cyclic, we have $\angle FAH = \angle O_{1}OO_{2}$. From this, together with (1) and (2) we get
$$\angle DLO = \angle O_{2}AH = \angle FAH - \angle FAO_{2} = \angle O_{1}OO_{2} - \angle POK = \angle UOP = \angle LOP$$
Therefore $OP$ is parallel to $\ell_{B}$ and $\ell_{C}$.