After filling with dielectrics the two capacitors will be in parallel order.
As shown, the two capacitors are connected in parallel. Initially the capacitance of capacitor
$ C = \frac{ε_0A}{d}$
where, $A$ is area of each plate and $d$ is the separation between the plates.
After filling with two dielectrics, we have two capacitors of capacitance.
$C_{1} = K_{1}\varepsilon_{0} \frac{\left(A/2\right)}{d}$
$ = \frac{4}{2} \frac{\varepsilon_{0}A}{d} $
$= \frac{4\varepsilon_{0}A}{d} = 4C $
and $C_{2} = \frac{K_{2}\varepsilon_{0}\left(A/2\right)}{d} $
$= \frac{4}{2} \frac{\varepsilon_{0}A}{d} $
$= \frac{2\varepsilon_{0}A}{d} = 2C$
Hence,their equivalent capacitance
$ C' = C_1+ C_2 = 4C + 2 C$
$= 6C$
i.e., new capacitance will be six times of the original.
