= Solution
In the fluid region $y>\eta(x)$, incompressibility and <potential flow> give
$$
\nabla^2\phi=0,\qquad \nabla\phi\to0\quad(y\to\infty).
$$
No penetration through the exact surface gives
$$
(-\eta_x,1)\mathbin\cdot(U+\phi_x,\phi_y)=0,\qquad
\phi_y=(U+\phi_x)\eta_x\quad(y=\eta).
$$
The condition $hk\ll1$ says the hill slope is small; $|\nabla\phi|\ll U$ says the disturbance <velocity> is small compared with the background wind. Dropping the product $\phi_x\eta_x$ and Taylor-shifting the boundary from $y=\eta$ to $y=0$ therefore gives
$$
\phi_y(x,0)=U\eta_x=-Uhk\sin kx.
$$
The decaying solution is
$$
\phi=Uh e^{-ky}\sin kx.
$$
Bernoulli's equation is
$$
p+\rho gy+\frac12\rho|Ue_x+\nabla\phi|^2=\text{constant}.
$$
To first order on the surface,
$$
p=\text{constant}-\rho g\eta-\rho U\phi_x,\qquad
\phi_x(x,0)=Uhk\cos kx.
$$
Thus
$$
p_{\rm trough}-p_{\rm crest}=2\rho h(g+kU^2).
$$
Equivalently, crest minus trough is the negative of this. For $kU^2/g\ll1$, hydrostatic elevation dominates; for $kU^2/g\gg1$, the Bernoulli <pressure> drop caused by faster crest flow dominates.
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