There are 2 possible cases – $P$ can be outside or inside of $\triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $\alpha = \beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $\theta = \alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $\phi = \theta$.