Your PFET will conduct all the time because it has a body diode that is forward biased in that orientation.
If you use an NFET in place of the PFET, as suggested in another answer, the circuit will "work", but the logic will be reversed. That is, the relay coil will be active when the gate is high, and the relay coil will be de-energized when they gate is low. However, you stated:
I'm working on a circuit that should close a reed relay (normally open) when TI/COUT is low and open said relay when TI/COUT is high (6V).
Spehro Pefhany beat me to the punch of getting a working CircuitLab simulation. But he has the right idea. The source pin of the PFET needs to be connected to the positive supply, making the switch a high-side switch.
One problem you may face with the particular PFET you are using is that its threshold voltage is only 0.4-1.0 V (per the CJ2301 datasheet). Therefore, to turn the relay off reliably, you need to bring the gate voltage to no less than the supply voltage minus 0.4 V. You might have better luck with a FET with a slightly larger threshold voltage.