In this case, if the volume is doubled and the number of molecules is doubled, while the temperature remains constant, the effects on pressure from both factors will cancel each other out. Doubling the volume would decrease the pressure by half, while doubling the number of molecules would increase the pressure by a factor of two. Therefore, the overall effect on pressure will be:
$$P_2 = P_1 \times \frac{V_1}{2V_2} \times \frac{2N_2}{N_1}$$
$$P_2 = P_1 \times 1 \times 1$$
$$P_2 = P_1$$
Therefore, the pressure of the gas will remain the same as the initial pressure.