= Solution
For the <second-order rough-surface scattered field>, the <Dirichlet boundary condition> expanded at the mean plane is
$$
\psi_s^{[2]}(x,0)=-h\partial_z\psi_s^{[1]}(x,0)-\frac{h^2}{2}\partial_z^2\psi^{[0]}(x,0).
$$
At normal incidence, $\psi^{[0]}=e^{-ikz}-e^{ikz}$, so its second normal derivative vanishes at zero. Define the <Dirichlet-to-Neumann map for a Helmholtz half-space> through the <Fourier multiplier> $\mathcal B$:
$$
\widehat{\mathcal B h}(q)=\beta(q)\widehat h(q),\qquad \partial_z\mathcal E g\big|_{z=0}=i\mathcal B g.
$$
The first-order trace is $2ikh$, hence $\partial_z\psi_s^{[1]}(x,0)=-2k\mathcal B h(x)$. It follows that
$$
\boxed{\psi_s(x,0)=-1+2ikh(x)+2k\,h(x)\mathcal B h(x)+O(\varepsilon^3),}
$$
where $h=O(\varepsilon)$ with fixed regular profile. In integral notation the quadratic contribution is
$$
2k\,h(x)\mathcal B h(x)=\frac{k}{\pi}h(x)\int\beta(q)\widehat h(q)e^{iqx}dq.
$$
The second-order field above the mean plane is $\mathcal E[2k h\mathcal B h]$ added to $-e^{ikz}+\mathcal E[2ikh]$. No local replacement of $\beta(q)$ by $k$ has been made; such a replacement would be an additional long-spatial-scale approximation.
The <physical surface trace and reference-plane trace> are distinct. At the actual rough boundary $z=h(x)$, the condition itself says $\psi_s(x,h(x))=-e^{-ikh(x)}$, so
$$
\boxed{\psi_s(x,h(x))=-1+ikh(x)+\frac{k^2h(x)^2}{2}+O(\varepsilon^3).}
$$
The first boxed expression is the reference-plane trace needed in part (d), at $z=0$, using the perturbative continuation where that plane lies below the actual boundary; the second answers the literal “at the surface” wording if it means the physical boundary. Taylor-expanding the first expression and its normal derivatives from $z=0$ to $z=h$ reproduces the second, so there is no contradiction between them.
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