Answer
The maximum number of lamps is $~~3$
Work Step by Step
We can find the resistance of each lamp:
$P = \frac{V^2}{R}$
$R = \frac{V^2}{P}$
$R = \frac{(120~V)^2}{500~W}$
$R = 28.8~\Omega$
We can find the equivalent resistance of $N$ lamps in parallel:
$R_{eq} = \frac{R}{N} = \frac{28.8~\Omega}{N}$
We can find the number of lamps which would result in a current of $15~A$:
$i = \frac{V}{R_{eq}}$
$i = \frac{V}{R/N}$
$i = \frac{V~N}{R}$
$i = \frac{(120~V)~N}{28.8~\Omega} = 15.0~A$
$N = \frac{(15.0~A)(28.8~\Omega)}{120~V}$
$N = 3.6$
Since a greater number of lamps would results in a higher current, the maximum number of lamps is $~~3$
