$(1)\quad\text{$T$ is complete.}$
$(2)\quad\text{$T$ is consistent and for any two models $M$,$N$ of $T$, $M\equiv N$.}$
$(3)\quad\text{There is a model $M$ of $T$ such that $C_n(T)=Th(M)$, where $C_n(T)=\{\sigma\in\mathcal L\;|\;T\vDash\sigma\}$.}$
$(4)\quad\text{$T$ has a model and for every model $M$ of $T$, $C_n(T)=Th(M)$.}$
I've shown that $(1)\iff(2)$
$(\implies)$
Suppose that $T$ is complete. Then, $T$ is consistent, and for all $\sigma\in Sent(\mathcal L),\;T\vDash\sigma\;$ or $\;T\vDash\lnot\sigma.$
Let $M,N\vDash T$. Clearly, $C_n(T)=Th(M)$ since $T$ is complete. ($C_n(T)$:maximal consistent in $\mathcal L$ and $M\vDash T$.)
Similarly, $Th(N)=C_n(T)$. Hence, $Th(M)=Th(N)$.
$(\impliedby)$
Assume that $(2)$ holds. Then, suppose that $T$ is not complete.
Then, there is $\sigma$ such that $T\not\vDash\sigma$, $T\not\vDash\lnot\sigma$).
So, both $T\cup\{\sigma\}$ and $T\cup\{\lnot\sigma\}$ are consistent.
But then since there are $M$,$N$ such that $M\vDash T\cup\{\sigma\}$ and $N\vDash T\cup\{\lnot\sigma\}$.
Clearly, $M\not\equiv N$. It's a contradiction. Hence, $T$ is complete.
Now I'm trying to prove the equivalence between $(3)$ and $(4)$.
Question: On the other hand is trivial. But on the one hand, I cannot approach appropriately.
And finally, I want to show $(1)$ is equivalent to $(3)$, by definition of $C_n(T)$.
$(3)\Rightarrow (4)$, For every $\phi\in \mathcal{L}$, $M\models \phi$ or $M\models \neg \phi$, so for every $\phi\in \mathcal{L}$, $\phi\in C_n(T)$ or $\neg\phi\in C_n(T)$. By $(3)$ $T$ has model. If $M'\models T$, then $C_n(T)\subseteq Th(M')$. If $\phi\in Th(M')$, because $T$ has a model and for every $\phi\in \mathcal{L}$, $\phi\in C_n(T)$ or $\neg\phi\in C_n(T)$, therefore $\phi\in C_n(T)$, so $Th(M')=C_n(T)$.