The reaction $$\ce{A.3H2O(s) <=> A.H2O(s) + 2H2O (g)}$$ has a $K_p = \pu{400 mmHg2}$
One millimole each of $\ce{A.3H2O}$ and $\ce{A.H2O}$ are taken in a $\pu{1 l}$ container at $\pu{300 K}$
Relative humidity is $80\ \%$ and saturation water vapor pressure is $\pu{85 mmHg}$ at $\pu{300 K}$. Find the final composition of the container
My attempt
\begin{align} K_p &=(P_\ce{H2O})^2 = 400 \\ \therefore P_\ce{H2O} &= \pu{20 mmHg} \end{align}
Now,
\begin{array}{lc} \hline \ce{&A.3H2O &<=> &A.H2O &+ &2H2O(g)} \\ \hline i &1\times10^{-3} && 1 \times10^{-3} && \pu{68 mmHg} \\ c &+x && -x && -P \\ e &2\times10^{-3} && 0 && \pu{30 mmHg} \\ \hline \end{array}
$\pu{38 mmHg}$ drops because $\pu{2 mmol}$ of water vapor was consumed
$$ P=\frac{1}{12}\frac{300\times 2\times 10^{-3}}{1} = \pu{38 mmHg}$$
The answer should have been $\pu{20 mmHg}$ pressure of water vapor but I am afraid there isn't enough reactant to make that much vapor pressure. I asked a friend and he suggested that we can take $\pu{2 mmol}$ each of the reactants as they are solid and don't influence the $K_p$ expressions while also not influencing the ratio of the added moles.
I do not get why this can be done but this allows for the vapor pressure at equilibrium to become $\pu{20 mmHg}$ as required