Set Membership and Set Definition Correct me if I am wrong, but I have a strong suspicion that the following two expressions are true. Again, this is only a suspicion. If somebody can point me wrong or right I will be satisfied. And sorry if the question is dumb.
\begin{equation}
a \in X \quad \text{ iff } \quad \exists x : (x \in X) \land (a = x) \tag{1}
\end{equation}
where $X$ is a finite set and $a$ a urelement.
\begin{equation}
\{x \mid x \in X \;\; \land \;\; P(x) \} \quad = \quad \{ x \mid \forall x \in X: P(x) \} \tag{2}
\end{equation}
where $X$ is a finite set and $P$ is a predicate with argument $x$.

When you define a set $A$ as 
\begin{align}
 A = \{x \mid Q(x)\}
\end{align}
you mean that, for every $z$ in your domain (possibly the universe if you are in naïve set theory), $z \in A$ if and only if $Q(z)$ holds, where $Q(z)$ is obtained from $Q(x)$ by replacing the free occurrences of $x$ with $z$.
Now, let $A = \{x \mid \forall x \in X : P(x)\}$ (the set on the right-hand side of identity $(2)$ in your question). This means that, for every $z$ in your domain, $z \in A$ if and only if $\forall x (x \in X \to P(x))$ holds ($z$ does not occur here because there is no free occurrence of $x$ where $z$ can substitute for it).
Therefore, $A$ is either the empty set (if $\forall x (x \in X \to P(x))$ is false) or the whole domain (if $\forall x (x \in X \to P(x))$ is true).
Clearly, the set $\{x \mid x \in X \land P(x)\}$ (the set on the left-hand side of identity $(2)$ in your question) can be different from the empty set and the whole domain. For instance, think of $X$ as the set of natural numbers and of $P(x)$ as "$x$ is even": in this case, $\{x \mid x \in X \land P(x)\}$ is the set of even natural numbers, in the domain of natural numbers.
So, your identity $(2)$ is wrong, it does not hold because it identifies two sets that are different in general. To make it true, you should remove the universal quantifier $\forall$ (because it binds the occurrences of $x$) from the set $\{x \mid \forall x \in X :P(x)\}$ in $(2)$.
In this way, we set $A = \{x \mid x \in X \land P(x)\}$ and hence, for every $z$ in your domain, 
\begin{align}
z \in A \ \text{  if and only if } \ z \in X \land P(z).
\end{align} 

Independently of from above, your equivalence $(1)$ is always true. More in general, for every property $P(x)$ (in your equivalence $(1)$, $P(x)$ is $x \in X$), the following holds:
\begin{align}
P(a) \ \text{ if and only if } \ \exists x \, (P(x) \land a = x).
\end{align}
