#1: To answer your question, yes, the drain of the MOSFET becomes floating/open-circuit (except for the body diode clamping to the supply). But since the LEDs aren't inductive loads, who cares? A floating node cannot provide current, current is required for the LEDs to light, ergo, making the node be floating will certainly turn the LEDs off. Instantly. The 10k resistors to ground just waste power.
#2: The brightness adjustment pot is a really dodgy idea, as depending on how many segments are on in your current row (which could be anywhere from 0 to
, the amount of current flowing through the pot will vary by a factor of 8. This means the drop across in varies by a factor of 8 as well. In summary, any significant resistance on that pot will have a minimal impact on the brightness of a single segment, but a severe impact on the brightness of an all-on display. It would be much preferable to just modify the duty cycle on the driving waveform instead, or have global blanking.
#3: The right-hand half of your diagram is horribly upside-down. The supply should always be at the top, ground at the bottom, and data flowing left to right. You've got all of those things backwards, which makes it very hard to read. P-type MOSFETs should always have their source terminals at the top. Put another way, those body diodes inside the MOSFETs should always be pointing upwards (as, generally speaking, they should never be carrying current except for weird spikes etc).