wow this is really cute and tricky. The condition $\angle BAD = 2\angle ADJ$ naturally leads us to try reflecting the line $DA$ over $DJ$, so that the resulting line is parallel to $AB$. Let this line intersect $\gamma$ for the second time at $R$. Note that $JP = JR = JQ$, so $J$ is the center of a circle passing through $P$, $Q$, and $R$. Moreover, this circle is tangent to $AB$ at $Q$ since $JQ \perp AB$. Let $QR$ intersect $\gamma$ at $S$. We claim that $AS$ is tangent to $\gamma$, and also $AS \perp DJ$, which will solve the problem. Claim 1. $A, Q, S, P$ are cyclic. Proof. Note that $D, R, S, P$ are cyclic. Since $DR \parallel AB$, therefore by Reim we have $DP \cap AB \equiv A, SR \cap AB \equiv Q, S, P$ are also concyclic. Claim 2. $AS$ is tangent to $\gamma$. Proof. Follows from alternate segment theorem by angle chase below: \[ \measuredangle ASP = \measuredangle AQP = \measuredangle QRP = \measuredangle SRP. \] Claim 3. $AS \perp DJ$. Proof. Follows from angle chasing below: \begin{align*} \angle PAS = \angle PQS &= \angle PQR \\ &= \frac{\angle PJR}{2} \\ &= 90^{\circ} - \frac{\angle PDR}{2} \\ &= 90^{\circ} - \angle PDJ, \end{align*}which means $AS \perp DJ$, as desired.

Happy to see this problem in a contest finally! I have 6 different solutions to this problem - initially I thought the problem is around medium level geometry instead, but if you know that its actually angle chasable, then the problem difficulty instantly decreases. (Commented this in the pdf below) I attach the pdf that I sent to the IGO PSC - it was initially rejected when I submitted it to the IMO 2024 (citing it was length bashable @.@), but fortunately this problem finally found a home in the end. [Also a very subtle JOM reference was intended - an inside joke for the Malaysians ]

Nice