The equilibrium constant for a chemical reaction is only a function of temperature. However, if you want to calculate the equilibrium constant $K$ via an equilibrium calculation, e.g. using the equation you posted, then we need to find the pressure dependence on the fugacity of the species $i$ as a liquid.
We write the relationship between the fugacity of a pure species $i$ with the Gibbs energy, for two different situations distinguished by a zero superscript
\begin{align}
g_i &= \Gamma(T) + RT\ln(f_i) \tag{1} \\
g_i^0 &= \Gamma(T) + RT\ln(f_i^0) \tag{2}
\end{align}
Subtracting Eq. (2) from Eq. (1)
$$ g_i - g_i^0 = RT\ln\bigg(\frac{f_i}{f_i^0}\bigg) \tag{3} $$
Equation (3) links two different states:
- The fugacity of a pure species at pressure $p$ and temperature $T$.
- The fugacity of a pure species at pressure $p^0$ and temperature $T$. The reference state for a pure liquid, in the context of this equation, is generally chosen as that where the substance at temperature $T$ is a saturated liquid, and thus $p^0 = p^\mathrm{sat}$.
Now we use the fundamental thermodynamic relation for pure substances
$$ \mathrm{d}g = v\mathrm{d}p - s\mathrm{d}T \tag{4} $$
The evolution $(p^\mathrm{sat},T)\to(p,T)$ at constant temperature, for a species labeled with $i$, is carried out by integrating Eq. (4)
$$ g_i - g_i^\mathrm{sat} = \int_{p_i^\mathrm{sat}}^p
v_i\mathrm{d}p \tag{5} $$
The liquid molar volume will change between these two pressures. For a liquid, we make the assumption that if the pressure range is not that large, it will remain constant and equal to the liquid saturated volume at the saturation pressure $ v_i = v_i^\mathrm{sat} $. Hence, Eq. (5) turns into
$$ g_i - g_i^\mathrm{sat} = v_i^\mathrm{sat}(p - p^\mathrm{sat}) \tag{6} $$
Combining Eqs. (3) and (6)
\begin{equation}
RT\ln\bigg(\frac{f_i}{f_i^\mathrm{sat}}\bigg) = v_i^\mathrm{sat}(p - p^\mathrm{sat})
\to
\frac{f_i}{f_i^\mathrm{sat}} = \exp\bigg[\frac{v_i^\mathrm{sat}
(p - p_i^\mathrm{sat})}{RT}\bigg] \tag{7}
\end{equation}
The right-hand side of Eq. (7) is the famous Poynting factor.
We return to the equilibrium constant. As stated, the mixture is ideal. We deal with liquids, so the fugacity coefficient of species $ i $ in a mixture is $\hat{f}_i = x_i\gamma_if_i= x_i f_i $, because $ \gamma_i = 1 $. Summarizing the reaction in the fashion $\ce{A + B -> C}$
\begin{equation}
K = \dfrac{\dfrac{\hat{f}_C}{f_C^\mathrm{sat}}}
{\left(\dfrac{\hat{f}_B}{f_B^\mathrm{sat}}\right)
\left(\dfrac{\hat{f}_A}{f_A^\mathrm{sat}}\right)} =
\dfrac{\dfrac{x_Cf_C}{f_C^\mathrm{sat}}}
{\left(\dfrac{x_Bf_B}{f_B^\mathrm{sat}}\right)
\left(\dfrac{x_Af_A}{f_A^\mathrm{sat}}\right)} \tag{8}
\end{equation}
Now we use Eq. (7) for the three ratios in Eq. (8)
\begin{equation}
\boxed{K =
\dfrac{x_C \exp\bigg[\dfrac{v_C^\mathrm{sat}
(\color{blue}{p} - p_C^\mathrm{sat})}{RT}\bigg]}
{x_A \exp\bigg[\dfrac{v_A^{sat}(\color{blue}{p} - p_A^{sat})}
{RT}\bigg] \cdot
x_B \exp\bigg[\dfrac{v_B^{sat}(\color{blue}{p} - p_B^{sat})}
{RT}\bigg]}} \tag{9}
\end{equation}
In blue is emphasized the pressure dependence in the fugacity.
Eq. (9) shows that additional information is needed when we deal with liquid-phase reactions:
- The saturation pressure $p_i^\mathrm{sat}$ for every species.
- The saturated liquid molar volumes $v_i^\mathrm{sat}$ for every species, evaluated at $p_i^\mathrm{sat}$.
