Let $R$ be the intersection point of the lines $(C Q)$ and $(A P)$. It suffices to prove the equality of the angles $(C B, C Q)=(A B, A R)$. To do this, it suffices to prove that the points $B, C, A, R$ are concyclic, that is, $(B R, B C)=(A R, R C)$. Since $(A R, A C) = (C P, C B)$, it suffices to prove that $(B R, B C)=(C P, C B)$, hence that the lines $(B R)$ and $(P C)$ are parallel.
For this last point, one can use that if $X=(C P) \cap(B R)$ then $(X, D, P, C)$ is a harmonic division. One can also do without projective geometry as follows:
We use Ceva's theorem in the triangle $P C E$, since the lines $(P B)$, $(D E)$, and $(R C)$ are concurrent at $Q$, with $R \in(P E)$ and $B \in(C E)$. We deduce that

$$1=\frac{C B}{E B} \cdot \frac{P D}{C D} \cdot \frac{R E}{R P}=\frac{C B}{E B} \cdot \frac{R E}{R P}$$

that is,

$$\frac{R E}{E B}=\frac{R P}{C B}$$

By applying the converse of Thales' theorem to the triangles $B R E$ and $C P E$, we deduce from this last equality that the lines $(B R)$ and $(P C)$ are indeed parallel, which concludes.
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