Three identical bulbs $B_1,B_2\text{\hspace{0.33em}and\hspace{0.33em}}{B}_3$ are connected as shown in the figure. Total power consumed by the circuit is ‘p’. If the bulb $B_2$ is fused, total power consumed by the system will be $P^1$. Then

If resistance of each bulb is R, then initial equivalent resistance of the system is, 

$R_C=\frac{R\times R}{R+R}+R=3R/2$.

Write $B_2$ gets fused, $R_C^1=R+R=2R$

Since power consumed by the system$=\frac{V^2}{R_C}$

We can write, $\raisebox{1ex}{$P^1$}\!\left/ \!\raisebox{-1ex}{$P$}\right.=\frac{R_C}{R_C^1}=\frac{3R/2}{2R}=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$

$\Rightarrow P^1=\raisebox{1ex}{$3P$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$