${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$

Pressure at equilibrium
${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{p}}_{{\mathrm{NO}}_2}^2}{{\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}=\frac{{\left(1.1\right)}^2}{0.28}=4.32 \mathrm{atm}$

If volume of the container is doubled, the pressure will reduced to half

${\mathrm{N}}_2{\mathrm{O}}_4\rightleftharpoons 2{\mathrm{N}}_2\mathrm{O}$

New pressure,

$\left[\frac{0.028}{2}-\mathrm{p}\right]\left[\frac{1.1}{2}+2\mathrm{p}\right]$

${\mathrm{K}}_{\mathrm{p}}=\frac{{\left[\frac{1.1}{2}+2\mathrm{p}\right]}^2}{\left[\frac{0.28}{2}-\mathrm{p}\right]}=4.32$

On solving, we get

$\mathrm{p}=0.045$

$\therefore {\mathrm{p}}_{{\mathrm{N}}_2{\mathrm{O}}_4}=0.14-0.045=0.095 \mathrm{atm}$

${\mathrm{p}}_{{\mathrm{NO}}_2}=0.55+\left(2\times 0.045\right)=0.64 \mathrm{atm}$