How is capacitance found when there are 2 different regions with dielectric in the in 2 parallel plate capacitors connected in series
If the diaphragm is metal (i.e. well-conducting), you can really look at the top capacitor independently from the bottom one: the electric field between the top conductor and the diaphragm only exists above the diaphragm.
You know the formula for capacitance of a plate capacitor:
$$C=\epsilon \frac Ad\text,$$
where \$A\$ is the area of the plates, \$d\$ is their distance, and \$\epsilon\$ the permittivity of the material between them.
Since a capacitor's plates are well conducting, you can just cut them at any point and model them as multiple smaller capacitors wired in parallel.
Now, cut the lower capacitor into two parts: one part with the blue dielectric in there, with an \$\epsilon_\text{blue}\$, and the part with the white dielectric, \$\epsilon_\text{white}\$.
This works because the electric field will be, in a pretty good approaximation, be exclusively between the plates, and exclusively perpendicular to the plates. So, there's no &quot;boundary effects&quot; between the blue and white dielectric.
From there, it's pretty clear that
$$ C_\text{total} = C_\text{blue}+C_\text{white} = \frac 1d\left(\epsilon_\text{blue}A_\text{blue} + \epsilon_\text{white}A_\text{white}\right)\text.$$
Assuming the blue dielectric has a permittivity that is very high compared to the white, i.e. \$\epsilon_\text{blue}\gg\epsilon_\text{white}\$,
$$C_\text{total}\approx \epsilon_\text{blue} \frac{A_\text{blue}}{d}\text;$$
meaning that the capacitance of the lower capacitor is a function of the area that the compressed dielectric covers, and of the distance. However, assuming there's a fixed volume of blue dielectric that gets squeezed, it follows geometrically that $$A_\text{blue}=\frac{V_\text{blue}}{d}\text.$$
Inserting that into the above formula yields  an inverse quadratic dependence of capacitance to distance, which yields an excellent sensitivity as a measurement device.