How to find the normalization constant of Fermi-Dirac distribution function? The Fermi-Dirac distribution function is given by 
$$f(E):=\dfrac{A}{{\mathrm e}^{(E-E_{F})\,/\,(k_{B}T)}+1},$$
where A is the normalization constant. When we sum over all the states, we get $1$. 
My confusion is why we sometimes write the distribution function as
$$f(E):=\frac{2}{{(2\pi \hbar)^{3}}}\dfrac{1}{{\mathrm e}^{(E-E_{F})\,/\,(k_{B}T)}+1},$$
where the  $2$ is due to two values of spin and ${(2\pi \hbar)^{3}}$ is a volume element in phase space.
 A: When normalized, $A$ is just equal to $1,$ so that $f(E)$ varies between $0<f(E)<1.$
Addendum for the edited question:
The prefactor $\frac{2}{(2\pi\hbar)^3}$ crops up in the volume integration of density of states performed in k-space for the computation of number of states $N$ (i.e. all available energy states up to a certain maximum (fermi level) are filled). Where the $2$ in the numerator is indeed to account for $2$ spin values, $(2\pi)^3$ in the denominator comes from the volume in k-space, and $\hbar^3$ appears when you rewrite the fermi wavenumber in terms of the fermi energy $k_F^3\propto \frac{(2mE_F)^{3/2}}{\hbar^3}.$
So the prefactor is not part of the fermi dirac distribution, actually another way to convince yourself would be to look at one of the many ways that the distribution can be derived. E.g. by maximizing the log of the multiplicity function $W$ (i.e. number of possible ways to fit the electrons in the number of available states) subject to two constraints, given for maximum number of electrons in the system and total energy:
$$
N=\sum_i g_if_i
$$
and 
$$
U=\sum_i E_i g_i f_i
$$
Using Lagrange multipliers to find the max of $\ln W$ you find the fermi dirac distribution. (for a thorough treatment see here). 
Alternatively, my favorite one is more in line with quantum statistics reasonings, it goes roughly as follows:
We consider the total grand canonical partition function: 
$$
Z=\sum_N \sum_i e^{-\beta(E_i-\mu_n)}
$$
Now in order to find the average occupation of some state $k$ we want to be able to express $Z$ as a sum of $Z_k$'s. The energy for $n_k$ particles occupying the state $\epsilon_k:$ $E_i=\sum_k n_k \epsilon_k.$
Thus 
$$
Z=\sum_k Z_k=\sum_k \sum_{n_k} e^{-n_k\beta(\epsilon_k-\mu)}
$$
From this the average occupation follows directly: (feel free to show this, and further expand the $\ln Z_k$, it's a useful exercise)
$$
\langle n_k \rangle = -\frac{1}{\beta}\frac{d}{d\epsilon_k}\ln Z_k
$$
Note that in the above we still had not assumed fermi statistics, we do so now: The many body wavefunction of fermions is antisymmetric $\psi(x_1,x_2,\cdots)=-\psi(x_2,x_1,\cdots)$ as no two fermions can occupy the same state, thus $n_k=0$ or $n_k=1.$ This greatly simplifies the computation of the partition function, $Z_k=1+e^{-\beta (\epsilon_k-\mu)},$ substituting back in $\langle n_k \rangle:$
$$
\langle n_k \rangle = \frac{1}{e^{\beta(\epsilon_k-\mu)}+1}
$$
which leads to the general expression of the fermi dirac distribution with the correct prefactor:
$$
f(E,T)=\frac{1}{e^{(E-E_f)/{k_B T}}+1}
$$
