It's not a debouncing circuit. All it does is briefly provide more current when the switch is first closed (for some reason). With 5 volt Vcc it would start with about 1/2 milliamp into the base of Q1, which would subside to 1/4 milliamp as long as the switch was closed. Both levels saturate Q1, and make OUT high, so long as GATE IN at J1 doesn't go negative by hundreds of volts.
If Vcc is the logical HIGH for the signal you want, you get that at OUT, but it's not debounced. Are you sure you need it to be?
If you meant "replace with a jumper" by "omit" R23, you won't be shorting Vcc to ground. You'll just be undoing whatever the circuit was intended to do. If you meant "remove" Elliot Alderson's answer is correct.
You need to provide more information if you want a useful answer. "What are you trying to do?" and "What's GATE IN?" lead the list of questions you should answer.