Which is somewhat surprising in itself.
I note your previous discussion in another area. Whilst this is not quite "cross-posting" as such, which is anathema to the moderation of these discussions, it is generally unproductive to on the one hand, ask for advice on different aspects of the same project in different places, and on the other, to fail to give essential information such as the circuit diagram you have just posted, in each and every such question.
Well, it isn't.
Hard to say, since you show no 100k resistors in the diagram. 
FAQ. A pull-up is almost always preferable since it avoids wiring the supply voltage itself out to pushbuttons, and especially if those pushbuttons are remote from the PCB, which always risks various consequences of an accidental short from the Vcc to ground - or to some other power source. Having the pushbuttons connect to ground (and therefore using a pull-up) means that the worst that can generally happen from a short circuit is a "cipher" with the button appearing to be pressed when it is not.
Well, since the buttons to which the 4051 (and we are I hope talking about a 74HC4051 rather than a CD4051) has not selected at any given moment, are entirely irrelevant at that time, there is no need to connect a pull-up to them. So if you use a pull-up on the common ("Z") terminal of the 4051, it will act on whichever button is currently selected - the only time you need it to.
The Arduino however provides a built-in pull-up function by setting "pinMode(2,INPUT_PULLUP);" (if in this case, 2 is the pin in question), so you generally need no pull-up at all - you use the Arduino itself.
And - there is no reason to use an analog input for this, though they are perfectly functional as digital I/O.
Question is, if it is called the "ground" on the 4051 pinout, why would you not - connect it to ground as specified? What makes you think a resistor should be placed between this ground terminal, and ground?
Only because you are using CMOS which has high impedances and negligible static current draw, does placing such a resistor here actually permit the circuit to function at all.
Now as to your previous questions about debouncing. In order to poll and debounce eight inputs, the proper way is on each pass through the master loop where you have determined that a new millisecond has since passed, to step through each index of the multiplexer, shifting in sequence the corresponding pushbutton state into successive bits of a byte value which then represents the combined value of the eight buttons. You then compare this value to the "previous" state which you have noted; if it is the same then you reset the debounce count variable to its maximum - say, 20 - and complete this section of code (back into the main loop). If the (byte) value just read is not the same, decrement the debounce counter and see whether it is zero or not. If it is not zero, this means that the value has not been stable for a full debounce period (20 milliseconds) so you simply complete this section of code. Finally, if the debounce counter is zero, then the current value clearly has been stable for a full debounce period so you then compare it to the "previous" or "last stable" state that you have noted (in a variable). Where it differs from the last stable state then you conclude that this is a new stable state which should be acted upon, so you examine and act upon each bit which differs from that last state, re-writing the "previous" state to match the bit acted upon until the "last" state now matches the current state and you move on from there.
I suppose I should code this algorithm into a generic reply to these questions. 
Note that there may be different possible valid sequences of the steps involved.
The importance of this is that it debounces eight (extensible to essentially any multiple of eight) inputs at a time, such as the code I wrote to service the keyboard of the Tandy TRS-80 "color computer" back in the 1980s. 
