4. Let $A$ and $B$ be the vertices of the major axis of the ellipse $\Gamma$, $E$ and $F$ be the two foci of $\Gamma$, $|A B|=4,|A F|=2+\sqrt{3}$, and $P$ be a point on $\Gamma$ such that $|P E||P F|=2$. Then the area of $\triangle P E F$ is $\qquad$

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4.1.

Let's assume the standard equation of the parabola $\Gamma$ in the Cartesian coordinate system is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$.
According to the conditions, we have $2 a=|A B|=4$,
$a \pm \sqrt{a^{2}-b^{2}}=|A F|=2+\sqrt{3}$.
Thus, $a=2, b=1$, and $|E F|=2 \sqrt{a^{2}-b^{2}}=2 \sqrt{3}$.
By the definition of an ellipse, we know $|P E|+|P F|=2 a=4$.
Also, $|P E||P F|=2$, so we get
$|P E|^{2}+|P F|^{2}$
$=(|P E|+|P F|)^{2}-2|P E||P F|$
$=12=|E F|^{2}$.
Therefore, $\angle E P F=90^{\circ}$.
Thus, $S_{\triangle P E F}=\frac{1}{2}|P E||P F|=1$.