BeanieBots is right, the same fundamental limitation on FET driving will exist if you us a pull-up or pull-down resistor. That being said, I'd try it with the P-Channel and pull-up resistors for a simple solution. It looks like you are driving LEDs; that's a low-inductance load, which makes it much more likely that you can get away with a bare-minimum FET driver. Why? Many FET failures happen when an inductive spike is coupled through a parasitic capacitance through the FET to the gate; if the gate is only being held by a weak pull-up, then the FET will switch when it shouldn't. This is also not really a problem here because there is no unlimited "shoot through" current as there is in a motor driver; current is always limited by the series resistor. If the FET accidently switched once in a while, it won't just fry.
There's a lot of good theory on FET driving, but all practical designs start with a decent topology and then swap parts until the FETs aren't running too hot or blowing up all the time. Use a precision thermometer (eg, your thumb) to monitor your FETs; if they heat up badly, most likely they aren't being turned off by the pull-up resistor fast enough (turn-off time much more critical than turn-on time, and your turn-off circuit is the weak link in this scheme). Go to a smaller pull-up or a FET with lower gate capacitance.
You'll want a small resistor in series with the gate also. Often people parallel a reversed diode with this resistor so that turn-on is limited but turn-off isn't (or is differently limited). But that's probably not necessary if your turn-off is driven by a pull-up and not a low-impedence driver.
Finally, a hardware question! I'll savor this day.
] goes to base. When base is grounded collector goes high... MOSFET goes on. Keep track of what voltages are allowed where and what the open switch potential is from ground...
