Equality of the formulae $S=k_B\ln\Omega(\bar E)$ and $S=-k_B \sum_i p_i\ln p_i$ for the canonical ensemble Reif's book on Statistical Physics is one of the most prescribed books in the subject. For the canonical ensemble, he derives that the formula for entropy is $$S_{\rm can}=k_B\ln\Omega(\bar E)$$ which differs from the microcanonical Boltzmann formula (or definition) $$S_{\rm mic}=k_B\ln\Omega(E)$$ i.e. $E$ has been changed to $\bar{E}$ in the argument of $\Omega$, the number of microstates of the system. Now, there is another expression for entropy for the canonical ensemble which is given by $$S_{\rm can}=-k_B \sum_i p_i\ln p_i,~~{\rm where}~~p_i=\frac{e^{-\beta E_i}}{\sum_j e^{-\beta E_j}}.$$


*

*I want to show that the two different looking formulae for $S_{\rm can}$ are identical by starting from any one of them and reduce it to the other.


Please see that my question is not the same as this. Note the argument of $\Omega$ in my formula (or refer to Reif's book). In no way, it is a duplicate. yu-v's answer seems correct.
 A: (I use $k_B=1$)
Reif gives, in $6.6.7$, the following:
$$ \ln(Z) = \ln \Omega(\bar{E}) - \beta \bar{E}$$
which he derives from $Z=\sum_E \Omega(E) \exp(-\beta E)$ and thermodynamics arguments.
So we have
$$ - \sum p_i \ln p_i = \frac{1}{Z}\sum_i \beta E_i e^{-\beta E_i} + \sum_i p_i \ln(Z) = \frac{\beta}{Z}\sum_i E_i e^{-\beta E_i}+\ln(Z) = \beta \bar{E}+\ln(Z) = \ln\Omega(\bar{E})$$
by definition. 
We can also further connect the sum over probabilities in general to the partition function and from it to the free energy
$$ - \sum p_i \ln p_i = \frac{1}{Z}\sum_i \beta E_i e^{-\beta E_i} + \sum_i p_i \ln(Z) = \frac{\beta}{Z}\sum_i E_i e^{-\beta E_i}+\ln(Z)=$$
$$ -\frac{\beta}{Z}\partial_\beta Z + \ln(Z) = -\beta \partial_\beta \ln(Z) + \ln(Z) = -\beta^2 \partial_\beta \frac{\ln(Z)}{\beta} = \partial_T T\ln(Z)$$
For the canonical ensemble $Z=\exp(-F/T)$ so we get $S_{can}=-\partial_T F$ which is consistent with $F=U-ST$.
A: I think that the comments/answers to the cited question completely answer yours:
In statistical physics the probabilities of all the microstates are assumed to be equal, that is 
$$p_i = \frac{1}{\Omega},$$
where Omega is the number of microstates (i.e. the number of terms in the sum for the entropy):
$$S = -k_B\sum_ip_i\log p_i = -k_B\sum_i\frac{1}{\Omega}\log \frac{1}{\Omega} =
-k_B\Omega \frac{1}{\Omega}(-\log \Omega) = k_B\log\Omega.$$
