Five identical capacitor plates each of area $A$ are arranged such that adjacent plates are at a distance $d$ apart. The plates are connected to a source of emf $V$ as shown in figure.  $\frac{-P{\epsilon }_{0}AV}{d}$ is the sum of charges on plate 1 and on plate 4, find value of $P$?

When the plates are alternately connected, the capacitors are treated in parallel. In the diagram we have four capacitors, so total capacitance is $\frac{4{\epsilon }_{0}A}{d}$. Now total charge on positive or negative plates is $\frac{4{\epsilon }_{0}AV}{d}$. Remember that these positive or negative charges are distributed in four layers as shown below. On the outer plates no layer has been shown because no charge is there to bind are on both the sides of these plates.

You know, that as per the conservation of charge principle, positive and negative charges will be equal in magnitude. Now, as there are four layers of charges and the total charge is $\frac{4{\epsilon }_{0}AV}{d};$  the charge on single layer is $\frac{{\epsilon }_{0}AV}{d}$. So, on plate 1, we have single layer positive charge equal to $\frac{{\epsilon }_{0}AV}{d}$ . On plate 4, we have double layer of negative charges, and its value is $\frac{-2{\epsilon }_{0}AV}{d}$ .