 Solution

For the absolute value function, $px-|x|\le 0$ for all $x$ if $|p|\le 1$, with equality at $x=0$. If $|p|>1$, taking $x=t\mathrm{sign}(p)$ with $t\to \infty$ makes the expression diverge. Thus its convex conjugate isThis is the indicator functional of a constraint set for the dual unit interval, not a zero-one indicator function.