A Buffer is created by mixing equal concentrations of NH4+ and NH3. Calculate the volume of both?

The easiest way to see this is to use the Henderson-Hasselbalch equation.  It states:

 ......................[NH3]
pH = pK + log --------------
.......................[NH4^+]

The equilibrium is:  NH4^+(aq) + H2O(l) = NH3(aq) + H3O^+(aq)

Ka = [H3O+][NH3]/[NH4^+] = 5.6 x 10^-10, so pKa = 9.25

According to the equation, if you have equal concentrations of NH3 and NH4^+ in a buffer solution the pH must be 9.25.  In order to produce a buffer with a pH of 8.75 the ratio of the components, NH3 and NH4^+ must not be equal.

So, mathematically,

8.75 = 9.25 + log [NH3]/[NH4^+]
log [NH3]/[NH4^+] = -0.50

[NH3]/[NH4^+] = antilog 10^-0.50 = 0.3162

So, the relationship between [NH3] and [NH4^+] in the buffer solution is:
[NH3] = 0.3162[NH4+]  This tells us that the [NH4^+] must be greater than the [NH3].

Without actually knowing the concentrations of the stock solutions containing NH4^+ and NH3 you can't really calculate the volume of each solution that must be added to obtain 100 mL of buffer solution required.

Your question implies that equal concentrations of NH4+ and NH3 in a buffer solution will give a pH of 8.75.  This is just not true.