detailed solution

Correct option is B

When switch is open :

X and Y lamps are in series. 

Power dissipated in the bulbs is

$P = i^{2}R$

and current i is 

$i = \frac{V}{2R}$

$\Rightarrow P_X=P_Y = \frac{V^2}{4R^2}R=\frac{V^2}{4R}$

When  switch is closed :

Y and Z are in parallel, their effective resistance is R/2. 

X and YZ combination are in series, and effective resistance becomes 3R/2

Current in the circuit is 

$i = \frac{2V}{3R}$

$\Rightarrow P_X=i^{2}R={(3V/2R)}^{2}R= \frac{9V}{}$

P.d across Y-Z combination is 

$V\text{'} = V-\frac{2V}{3R}R=\frac{V}{3}$

$\Rightarrow P_Y=P_Z = \frac{{(V/3)}^2}{R}=\frac{V^2}{9R}$

Therefore brightness of X increases and Y decreases.