= Solution
Because $\zeta<\eta\leq h(x)$, the lower layer is saturated along the entire barrier. Both layers have the same horizontal head gradient but different <Darcy velocities>. Integrating their conductivity over the actual cross-section gives the transmissivity
$$
\mathcal T(h)=\frac{\rho g}{\mu}\left[k_1\frac{\alpha\zeta^3}{3}
+k_2\frac{\alpha(h^3-\zeta^3)}{3}\right]
=\frac{\rho g\alpha}{3\mu}\left[k_2h^3+(k_1-k_2)\zeta^3\right].
$$
The total discharge is $Q=-\mathcal T(h)h_x$. Hence $QL=\int_\eta^H\mathcal T(h)dh$, giving
$$
\boxed{Q=\frac{\rho g\alpha}{3\mu L}
\left[\frac{k_2}{4}(H^4-\eta^4)+(k_1-k_2)\zeta^3(H-\eta)\right].}
$$
This is a parallel-layer conductivity integral, not a series-resistance average: the layers share the horizontal pressure drop. It reduces to the previous result when $k_1=k_2$. The assumption $\zeta<\eta$ is important; if the downstream free surface cuts the lower layer, the transmissivity and its head integration must be split into different saturation regimes.
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