Taiwan IMO Selection Camp

First, if $AB$ is parallel to $L$, draw the perpendicular bisector of $AB$ meeting $L$ at $R$, and draw the circumscribed circle $C$ of $\triangle ABP$. Then, for any point $P$ on $L$ other than $R$, let $PB$ meet circle $C$ at $S$, then $\angle APB < \angle ASB = \angle ARB$. Hence there exists a unique point $R$ satisfying the condition of the problem.

Next, if $AB$ is not parallel to $L$, let the line $AB$ meet the line $L$ at $K$. It is easy to see that there exist in the plane exactly two circles $C$ and $C'$, both passing through the points $A$ and $B$, and tangent to $L$. Let these two circles be tangent to $L$ at $R$ and $R'$ respectively.

Now, without loss of generality assume $\angle AR'B \le \angle ARB$, then for any point $P$ on the line $L$ other than $R$ and $R'$,
(i) if $P$ is on the same side as $R$, let $PB$ meet circle $C$ at $S$, then $\angle APB < \angle ASB = \angle ARB$.
(ii) if $P$ is on the same side as $R'$, let $PB$ meet circle $C'$ at $S'$, then $\angle APB < \angle AS'B = \angle AR'B \le \angle ARB$.

Therefore, there exists a unique point $R$ satisfying the condition of the problem, if and only if $\angle AR'B \ne \angle ARB$, if and only if $AB$ is not perpendicular to $L$.