4. Let $A, B$ be the vertices of the major axis of ellipse $\Gamma$, $E, F$ be the two foci of $\Gamma$, $|A B|=4,|A F|=2+\sqrt{3}$, $P$ be a point on $\Gamma$, satisfying $|P E| \cdot|P F|=2$, then the area of $\triangle P E F$ is $\qquad$
According to the problem, $a=2, a+c=2+\sqrt{3}$, then $c=\sqrt{3}, b=1$. Let $\angle E P F=\theta,|P E|=x,|P F|=y$, in $\triangle P E F$, by the cosine rule we get $4 c^{2}=x^{2}+y^{2}-2 x y \cos \theta=(x+y)^{2}-2 x y(1+\cos \theta)$.

Thus $12=16-4(1+\cos \theta) \Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}$, so $S_{\triangle P E F}=b^{2} \tan \frac{\theta}{2}=1$.