calculating Kp with partial pressures
I do agree with your point and Philipp explanation. Getting more formal, your $K_p$ at 35°C is:

$$ K^\circ_p(308.15\,\mathrm{K}) = \frac{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)^2\chi^2_\mathrm{NO_2}}{\left(\frac{p_\mathrm{tot}}{p_\circ}\right)\chi_\mathrm{N_2O_4}} $$

Where $p_\mathrm{tot}$ is the pressure in your flask and $p_\circ$ the standard pressure. The $\chi$ quantities stand for molar ratio. For the reaction given, those ratios are:

$$ \chi_\mathrm{NO_2} = \frac{n_{NO_2,0} + 2\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

and:

$$ \chi_\mathrm{N_2O_4} = \frac{n_{N_2O_4,0}  -\xi}{n_{N_2O_4,0} + n_{NO_2,0} + \xi} $$

Those identities come from amount balance, where $\xi$ is the molar coordinate of the reaction and zero-indexed quantities amount of matter when starting experience.

Assuming that initially there was only $\mathrm{N_2O_4}$, and you are using the following definition of dissociated ratio:

$$ \tau = \frac{\xi}{n_\mathrm{N_2O_4},0} = 0.15$$

You will obtain by substitution, molar ratio values:

$$ \chi_\mathrm{N_2O_4} = \frac{0.85}{1.15},  \chi_\mathrm{NO_2} = \frac{0.3}{1.15}$$

Then you inject those values in the $K^\circ_p$ expression above. Do not forget to express pressures in bars - this is a common mistake, you must remeber that $K_p$ are dimensionless. Therefore, $p_\mathrm{tot} = 1.2 \times 1.01325 \,\mathrm{bar}$ and $p_\circ = 1.0 \, \mathrm{bar}$. This is also an approximation because we are assimilating pressure as activities. But in your case you can consider fugacities close to unity and assimilate your gas mixture as an ideal gas.

Yes, you were right about molar ratio. Do not forget that equilibrium constants are dimensionless: they are ratio of activities!