= Solution
{c}
\b[Notation.] The source uses $p_s$ for both the hypothetical dry surface <pressure> and the actual surface <pressure> after <evaporation>. Distinguish them as $P_d$ and $P$, respectively, and interpret the specified <pressure> thickness $\Delta p$ as fixed. This conserves the original <hydrogen> inventory and permits $P$ to vary. Reading both occurrences as one fixed number would be inconsistent with adding <water vapor> to a closed <atmospheric column mass> budget.
Let $E(T)$ denote the <saturation pressure> of <water>, let $x$ be the <mole fraction> of <water vapor> in the mixed layer, and let $M_d,M_v$ be the <molar masses> of <hydrogen> and <water>. Surface <phase equilibrium> and the <ideal gas law> give
$$
x=\frac{E(T_s)}P,\qquad p_v(p)=xp.
$$
Because the layer has constant <temperature>, its <relative humidity> is
$$
\boxed{\mathrm{RH}(p)=\frac{p_v(p)}{E(T_s)}=\frac pP},\qquad P-\Delta p\leq p\leq P.
$$
Thus every level strictly above the surface is subsaturated: <hydrostatic balance> makes $p<P$, whereas the <saturation pressure> stays constant. The result does not require a dilute <water vapor> approximation.
The <Clausius-Clapeyron relation>, neglecting the <specific volume> of <liquid water> and treating the <water vapor> as an <ideal gas>, gives
$$
\frac{d\ln E}{dT}=\frac{L_v}{R_vT^2},\qquad
E(T)=E_t\exp\!\left[\frac{L_v}{R_v}\left(\frac1{T_t}-\frac1T\right)\right],\qquad R_v=\frac R{M_v}.
$$
Here $T_t,E_t$ are the <triple point> values, $L_v$ is the specific <latent heat>, and $R$ is the <molar gas constant>. All <temperatures> in these expressions are absolute. As throughout the question, this is a constant-<latent heat>, <ideal gas> model, not an extrapolation valid through the real <critical point> of <water>.
For a well-mixed <ideal gas> mixture, its mean <molar mass> and <water vapor> <mass fraction> are
$$
\overline M=M_d(1+ax),\qquad w_v=\frac{\epsilon x}{1+ax},\qquad
\epsilon=\frac{M_v}{M_d}\simeq9,\qquad a=\epsilon-1\simeq8.
$$
By <hydrostatic balance>, the mixed layer has total <mass> per unit area $\Delta p/g$, of which $w_v\Delta p/g$ is <water vapor>. The remaining <atmosphere> contains no <water vapor>. Consequently conservation of <hydrogen> <mass> gives
$$
\frac P g=\frac{P_d}g+\frac{\Delta p}g w_v,
\qquad
P=P_d+\Delta p\frac{\epsilon E(T_s)}{P+aE(T_s)}.
$$
Equivalently the requested <quadratic equation> is
$$
\boxed{(P-P_d)(P+aE)=\epsilon\Delta p\,E.}
$$
For each $T_s$, calculate $E$, take the positive <root> of this <quadratic equation>, and retain it only if $0\leq x<1$ and $P>\Delta p$. The last condition leaves a nonempty dry layer. In particular the dry constituent's surface <partial pressure> is not generally $P_d$: $P_d$ measures its column weight, and the <hydrogen> and <water vapor> have different <molar masses>.
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