I don't know for sure, but... the description "power supply decoupling" implies that a decoupling capacitor could/should be connected between that pin and ground. A decoupling capacitor provides surges of energy when the chip requires it, that if there were no capacitor there, would need to come over a long copper trace to the power supply and would suffer an instantaneous voltage drop. You COULD connect it directly to the power supply bus, but I think the recommendation would be to connect a capacitor.
You should look for an example circuit. Probably, internal to the chip, that pin is also connected to the Vdd pin, so that you just need a capacitor. But maybe example circuits also connect that pin to the power bus (so there is a network there with three connections: DEC2, + pin of capacitor, power bus.)
If it is connected internally, the chip is providing (and consuming) voltage on the pin the same as on the power bus. You should NOT connect it to a different power bus with a higher voltage than is connected to the Vdd pin.
Same for DEC1. In other words, pins Vdd, DEC1, and DEC2 are all just externally available nodes on the internal network of the chip that distributes power inside the chip, and should be connected to the same external power bus, with decoupling capacitors close by (on at least one of the pins.)
It could be that there are many pins just for convenience of laying out a decoupling capacitor on at least one of them. Again, you probably should study example circuits, or find a recommendation from Nordic.
DEC2 is for decoupling the 1.7V LDO regulator in normal mode. In low voltage mode you can bypass the regulator and apply 1.75 - 1.95 V. 3V will probably break the chip.
