Solution

The pressure inside the droplet is the sum of the Laplace pressure and the pressure of the surrounding liquid. Let $x$ be the distance from the liquid surface to the point inside the droplet,then the pressure inside the droplet:

\begin{equation} P(x)=\rho gx+\frac{2\sigma}{R} \end{equation}

Now we see that the resulting function is linear,so we get that the pressure reaches the minimum value when $x=h-R$ and maximum value when $x=h+R$,so:

\begin{equation} P_{max}=\rho g(h+R)+\frac{2\sigma}{R} \end{equation}

\begin{equation} P_{min}=\rho g(h-R)+\frac{2\sigma}{R} \end{equation}