With this description:
Similar to the NPN circuit, the base is our input, and the emitter is tied to a constant voltage. This time however, the emitter is tied high, and the load is connected to the transistor on the ground side.
This circuit works just as well as the NPN-based switch, but there’s one huge difference: to turn the load “on” the base must be low. This can cause complications, especially if the load’s high voltage (VCC in this picture) is higher than our control input’s high voltage. For example, this circuit wouldn’t work if you were trying to use a 5V-operating Arduino to switch on a 12V motor. In that case it’d be impossible to turn the switch off because VB would always be less than VE.
I am confused because the part about the Arduino not being able to turn off a PNP transistor makes sense, but doesn't the circuit shown feature exactly the situation described that the transistor wouldn't work in? Is this an oversight by me and is there something else going on here that I don't understand, or is this a mistake in the article?