Answer
See explanation
Work Step by Step
The definition for enthalpy is $$
h=u+P v
$$ Then, $$
\left(\frac{\partial h}{\partial P}\right)_T=\left(\frac{\partial u}{\partial P}\right)_T+P\left(\frac{\partial v}{\partial P}\right)_T+v\left(\frac{\partial P}{\partial P}\right)_T
$$ Assume $u=u(s, v)$
Then, $$
\begin{aligned}
& d u=\left(\frac{\partial u}{\partial s}\right)_v d s+\left(\frac{\partial u}{\partial v}\right)_s d v \\
& \left(\frac{\partial u}{\partial P}\right)_T=\left(\frac{\partial u}{\partial s}\right)_v\left(\frac{\partial s}{\partial P}\right)_T+\left(\frac{\partial u}{\partial v}\right)_x\left(\frac{\partial v}{\partial P}\right)_T \\
& \left(\frac{\partial u}{\partial P}\right)_T=T\left[-\left(\frac{\partial v}{\partial T}\right)_P\right]-P\left(\frac{\partial v}{\partial P}\right)_T=-(T+P)\left(\frac{\partial v}{\partial P}\right)_T \\
& \left(\frac{\partial h}{\partial P}\right)_T=-(T+P)\left(\frac{\partial v}{\partial P}\right)_T+P\left(\frac{\partial v}{\partial P}\right)_T+v=-T\left(\frac{\partial v}{\partial P}\right)_T+v
\end{aligned}
$$ For ideal gases $$
v=\frac{R T}{P} \text { and }\left(\frac{\partial v}{\partial P}\right)_T=\frac{R}{P}
$$ Then, $$
\left(\frac{\partial h}{\partial P}\right)_T=-\frac{T R}{P}+v=-v+v=0
$$
