here is my solution:

we now the partial pressure of NO2 in equilibrium. Assuming the gasses behave as ideal gasses, p(NO2) is given by:

pNO2 = nRT/V

--> n(nO2) in equilibrium = (1,41855 x 10^5 * 1,0 x 10^-3) / (8,314 * 308) = 0,05540 mol

We started with 0,25 mol N2O4 so 0,05540/2 = 0,02770 mol N2O4 has reacted.

now we thus have, in equilibrium, 0,25 - 0,02770 = 0,2230 mol N2O4

--> p(N2O4) = 0,2230 * 8,314 * 308 / 1,0 x 10^-3 = 5,711 x 10^5 Pa.

This looks nice, since the Kp for this equilibrium at T = 298 K is 1,48 x 10^4 Pa.

Now Kp = p(NO2)^2 / p(N2O4)

--> Kp = (1,41855 x 10^5)^2 / 5,711 x 10^5 = 3,5 x 10^4 Pa.

( In this case, the relation between Kp and Kc:

Kp = Kc * RT

--> delta n = 2-1 = 1)