High-Voltage Switching Circuit – 48V Typical

There is no need for a opto-coupler here. In your circuit, it is not performing isolation since both sides are tied to the same ground. It is therefore just acting like a slow-responding transistor with very limited gain. Replace U2 with a NPN transistor or "logic level" N channel MOSFET.

Personally, I'd use the NPN transistor in this case. Just about any small signal NPN will do. I use 2N4401 (actually MMBT4401, the SOT-23 version, but that's more cumbersome to write) for such jellybean applications, but many many others would work fine too. Connect the emitter to ground, the base to the right side of R1, and the collector to the FET gate. Yes, it really is that simple.

Since a bare transistor will have more gain than the opto-isolator, you can increase R1. 1 mA base drive is more than enough in this case. Let's say the B-E drop of the NPN transistor is 700 mV. That leaves 2.6 V accross R1 when the transistor is supposed to be on. (2.6 V)/(1 mA) = 2.6 kΩ. That will be slower to turn off, but the FET turnoff time will be dominated by R2 acting against the FET's gate capacitance anyway.

This isn't a high speed switching application, right? If it were, you would need to make the FET turn off more deliberately than with just a 10 kΩ pullup. Your circuit is fine if the FET is just occasionally turning power on and off to some other circuit or device, not switching at more than 100 Hz or so.

Added:

I just noticed that you want the 12 V power to be on when the FPGA is "unconfigured", whatever that really means. I'll take it to mean that the P1 output will be floating at that time. In that case add a pullup resistor to the P1 output. With the NPN transistor as describe above, you don't really need much current thru its base to turn on the FET. Even a 10 kΩ pullup would be enough to keep the NPN transistor on, but high enough to not cause significant current when the FPGA is actively driving the line low to shut off the 12 V supply.

Here is the overall solution I am proposing:

Yeah, I think Option 2 is your simplest high-side option. Replace Q1 with a BSS138, and you can have a simple all MOSFET solution with discrete parts.

Alternately, if you're OK buying ICs, you can use an IC like the TPS2557 (which is an N-channel load switch with built-in charge pump), or an integrated dual N/P part like the Si3865DDV which basically squeezes the option 2 circuit into a convenient package. These would also take up less space than the discrete solution, and offer some protection features such as current-limiting if you would like. I can also understand the appeal of building something with parts you have lying around.

A N-channel device will have a lower on-resistance than a comparable P-channel device, but it does require a charge pump as you stated.

Finally, even with a MOSFET solution, I would personally keep R2 there and make it 1k or so, in addition to adding a pull-down to your UC enable pin so that when you are in reset, the LEDs stay off. By keeping R2, if you drop a screwdriver or something across the gate of the MOSFET to GND, it limits the current that can flow from the uC pin.

Here's the discrete solution:

^{simulate this circuit – Schematic created using CircuitLab}