Question: two steps

1) suppose $b$ is definable in $\mathcal{M} $ over $A$, then we need to show $b$ is definable over dcl$(A)$.

2) suppose $b$ is definable in $\mathcal{M}$ over dcl$(A)$, then we need to show $b$ is definable over $A$.

I think that we need to construct a formula which satisfy 1) in step 1), and construct a another formula which satisfy 2) in step 2).

But I really don't know how to construct the formulas. So I need your help. 

Thanks!

Answer: 
First, it should be obvious that $\mathrm{dcl}(A) \subseteq \mathrm{dcl}(\mathrm{dcl}(A))$. We must now show that $\mathrm{dcl}(\mathrm{dcl}(A)) \subseteq \mathrm{dcl}(A)$. 

Assume that $b \in \mathrm{dcl}(\mathrm{dcl}(A))$. Then $\exists \varphi(x,a\_1,...,a\_n)$ with $\varphi(M,a\_1,...,a\_n) = b$ and $a\_1,...,a\_n \in \mathrm{dcl}(A)$. Since $a\_1,...,a\_n \in \mathrm{dcl}(A) \implies \exists\psi\_i(x,\bar{c}\_i)$ such that $\bar{c}\_i \in A^{i}$ and $\psi\_i(M,\bar{c}\_i) = a\_i$

But then, consider the formula $\theta(x) = (\exists x\_1,..,x\_n)\bigwedge\_{i=1}^n \psi\_i(x\_i,\bar{c}\_i)\wedge\varphi(x,x\_1,...,x\_n)$. We now have that $\theta(M) = {b}$ and $\theta(x)$ contains parameters only from $A$. Therefore, $b \in \mathrm{dcl}(A)$ and we are done.
For a more geometric proof: notice that if $N\succeq M$, then for any $A\subseteq M$ we have that $\textrm {dcl}(A)$ remains the same in $N$ and $M$.

If you take $N$ to be a sufficiently homogeneous model, then whenever $A$ is a small set, $\operatorname{dcl}(A)$ is just the set of points which are fixed by all automorphisms of $N$ fixing $A$. Thus $\operatorname{dcl}(\operatorname{dcl}(A))$ is simply the set of all points which are fixed by all automorphisms of $N$ fixing $\operatorname{dcl}(A)$. But those are, by definition, exactly the automorphisms fixing $A$.