Let height of water in tank be $h$
$\text { So, } 4 P-P=\rho_w g h \quad \ldots (1)$
$\because \frac{3}{5}$ water taken out $\frac{2}{5}$ th water is left to exert pressure
$P^{\prime}=P+\frac{2}{5} \rho_w g h$
$\Rightarrow P^{\prime} =P+\frac{2}{5} \times 3 P$ [From eq. $(1)$]
$\Rightarrow P^{\prime} =\frac{11 P}{5}$
