The equilibrium constant $\left({\mathrm{K}}_{\mathrm{p}}\right)$ for the reaction, 

$2{\mathrm{SO}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(g\right)$ at 1000 K is 3.5 ${\mathrm{atm}}^{-1}$
What would be the partial pressure of oxygen gas, if the equilibrium is found to have equal moles of ${\mathrm{SO}}_2$ and ${\mathrm{SO}}_3$?

In terms of partial pressure, the equilibrium constant is ${\mathrm{K}}_{\mathrm{p}}$ for the reaction,

$\mathrm{aA}+\mathrm{bB}\rightleftharpoons \mathrm{cC}+\mathrm{dD}$

${\mathrm{K}}_{\mathrm{\rho }}=\frac{{\mathrm{P}}_{\mathrm{C}}^{\mathrm{c}}\times {\mathrm{P}}_{\mathrm{D}}^{\mathrm{d}}}{{\mathrm{P}}_{\mathrm{A}}^{\mathrm{a}}\times {\mathrm{P}}_{\mathrm{B}}^{\mathrm{b}}}$

Then for the given reaction,

$2{\mathrm{SO}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)\rightleftharpoons 2{\mathrm{SO}}_3\left(g\right)$, ${\mathrm{K}}_{\mathrm{p}}=3.5 {\mathrm{atm}}^{-1}$

So,

$\begin{array}{l}{\mathrm{K}}_{\mathrm{p}}=\frac{{\left({\mathrm{P}}_{{\mathrm{SO}}_3}\right)}^2}{{\left({\mathrm{P}}_{{\mathrm{SO}}_2}\right)}^2\left({\mathrm{P}}_{{\mathrm{O}}_2}\right)} [\because {\mathrm{P}}_{{\mathrm{SO}}_3}={\mathrm{P}}_{{\mathrm{SO}}_2}]\\ \therefore 3.5=\frac{1}{{\mathrm{P}}_{{\mathrm{O}}_2}}\\ {\mathrm{P}}_{{\mathrm{O}}_2}=0.285\end{array}$

So, option 4 is correct.