Electronic – Pulling a pin to 12 V using 3.3V

I've solved problems like that in the past with a Sziklai pair. It's like a Darlington, but with two complementary transistors. In your case, use a PNP transistor with its emitter connected to the alarm chip's VCC, and the base, through a resistor, to the open drain alarm output. The collector goes to the base of an NPN transistor (again, through a bias resistor) wired as a low-side switch for your 9 volt circuit.

A single transistor will not provide an interface with your FET AND do it without inversion. You can do the job with two transistors, though.

^{simulate this circuit – Schematic created using CircuitLab}

This will provide a decent drive for your FET with about a 1 usec added to on-time. That is, if you put in a 10 usec pulse, the FET will be on for about 11 usec.

Also note that R5 needs to be a 1-watt resistor.

EDIT -

It turns out I was using the wrong FET model, and my value for gate capacitance was too low. For a "real" IRFZ44, the added pulse width is about 3 usec, not 1. The problem is that, when you go to turn off the FET, The charge stored in the gate can only discharge through R5, and the time constant is in the 3 usec range. You can, of course, decrease R5, but then the power it dissipates when the load is driven on goes up. You can get a 2 usec delay rather than 3 by decreasing R5 to 100 ohms, but then you need a 2-watt resistor. If you need very narrow PWM cycles (low effective motor current) you're probably better off with a more sophisticated driver. I personally tend to go with the Maxim MAX4426/4427 (less than $4 each, with 2 channels per IC), but that's just habit.