The dissociation of $\ce{PCl5}$ is given by the equation:
$$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)}$$
with $\displaystyle K_\text{p} = \frac{p_{\ce{Cl2}}p_{\ce{PCl3}}}{p_{\ce{PCl5}}}$. Putting this in terms of total pressure, $\displaystyle K_\text{p} = \frac{x_{\ce{Cl2}}x_{\ce{PCl3}}}{x_{\ce{PCl5}}}p_\text{T}$.
In terms of the degree of dissociation, $\alpha$, assuming an initial amount $y$ of moles of $\ce{PCl5}$, at equilibrium, we have $n_{\ce{PCl5}} = y - \alpha y\,\text{mol}$, $n_{\ce{PCl3}} = \alpha y\,\text{mol}$, and $n_{\ce{Cl2}} = \alpha y\,\text{mol}$. Thus, their molar fractions are $x_{\ce{PCl5}} = \frac{1 - \alpha}{1 + \alpha}$, $x_{\ce{PCl3}} = \frac{\alpha}{1 + \alpha}$, and $x_{\ce{Cl2}} = \frac{\alpha}{1 + \alpha}$, so:
$$K_\text{p} = \frac{\left(\frac{\alpha}{1 + \alpha}\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right)}p_\text{T} = \frac{\alpha^2}{1-\alpha^2}p_\text{T}$$
Since at $p_\text{T} = \pu{2 atm}$, $\alpha = 0.4$, we get $K_\text{p} = \pu{0.381 atm}$. Hence, for $\alpha = 0.6$, since $K_\text{p}$ doesn't change (i.e. assuming it's an isothermic system), we get $p_\text{T} = \pu{0.677 atm}$. This is in agreement with Le Chatelier's principle: reducing the total pressure will push the equilibrium towards the side with more gaseous moles, so the dissociation constant in this case will be higher.
