$f(x,y) = x^{n}y^{m}=P$
Now, $x$ is doubled so we substitute $2x$ for $x$ AND $y$ is halved so we substitute $y/2$ for $y$
$f(x,y) = (2x)^{n}(\frac{y}{2})^{m}$
We get,
$f(x,y) = (2)^{n} (x)^{n}(y)^{m}(2)^{-m}$
$f(x,y) = 2^{n-m} x^{n}y^{m} = 2^{n-m} P$
Answer A) $2^{n-m}P$