Explanation

(b) Let the resistance of the two heaters be denoted by $R_{1}\ and\ R_{2}.$
Then, $R_{1} = \frac{V^{2}}{P_{1}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ R_{2} = \frac{V^{2}}{P_{2}}$
In parallel combination of resistances,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{R_{P}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{P_{P}}{V^{2}} = \frac{P_{1}}{V^{2}} + \frac{P_{2}}{V^{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{p} = P_{1} + P_{2}$
Given, $P_{1} = 1000\ W$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ P_{2} = 1000\ W$
$\therefore\ \ \ \ \ \ \ \ \ \ P_{p} = 1000 + 1000$
Hence, $P_{p} = 2000\ W$