Power supply for a device
The specifications say:

Power supply: 1.5 V AC (Max sine)
Operating frequency: 500 Hz - 2 kHz
Rated power: 0.2 mW (Max sine)
Central value: 31 kΩ (at 25 Celsius, 1 kHz ,1 V AC, 60% RH)

I would take the first to be 1.5 V peak and the second to be RMS. Given that a 1 Vrms sine would peak at \$ \sqrt 2\$ times the RMS value that would make sense.
As a second check look at their power rating for the central value - it's 31 kΩ at 1 V (RMS, we assume). From \$ P = \frac {V^2} R = \frac {1^2}{31k} = 30 \ \mu\text {W} \$ all would be fine.

Figure 1. Resistance vs RH.
Worst case will be at minimum resistance. At 90% RH the resistance falls to 2 kΩ. Calculating again we get \$ P = \frac {V^2} R = \frac {1^2}{2k} = 500 \ \mu\text {W} \$. At this point you are 2.5 times the rated power. Not so good.
I'd use the max rated power figure, the resistance for your worst-case %RH and work out your Vrms from that.