Dynamic equilibrium of gases
For an ideal gas mixture the total pressure is the sum of the partial pressures of all the gases in the mixture. To make intuitive sense of this, consider that since it is a gas mixture, all the gases essentially have access to the whole container, and so will homogenously fill it. Each gas wil the exert a certain pressure on the container walls. The sum the pressures from all the contributing gases will then equal the total pressure the container walls experience.

We also have the relationship $$p_\text{x}/p_\text{total}=n_\text{x}/n_\text{total}$$

The potential yield for the rxn is 2 mol ($\ce{SO2}$ limiting reactant). To get a 80% yield, we therefore have 1.6 mol $\ce{SO3}$ at equilibrium. Therefore there will be 0.4 mol $\ce{SO2}$ and 1.2 mol $\ce{O2}$ left at equilibrium. $n_\text{total}=3.2$

$n_\ce{SO2}/n_\text{total}=0.4/3.2=0.125$

$n_\ce{SO3}/n_\text{total}=1.6/3.2=0.5$

$n_\ce{O2}/n_\text{total}=1.2/3.2=0.375$

To find the partial pressures of the individual gases, we combine the ratios above with relationship 1

$p_\ce{SO2}=0.125\times p_\text{total}$
$p_\ce{SO3}=0.5\times p_\text{total}$
$p_\ce{O2}=0.375\times p_\text{total}$

As you know, the equilibrium can be expressed with partial pressures, and is constant for all changes in pressure. $K_\mathrm{p}=p_\ce{SO_3}^2/p_\ce{SO_2}^2p_\ce{O_2}=0.13$  at 830 Celsius

Plugging in the partial pressures and solving for p(total):

$(0.5x)^2/((0.125x)^2\times0.375x)=0.13$
$\implies x=328.205$ (The unit can be given by the units of the eq. constant.)