4. As shown in Figure 3, in $\triangle A B C$, it is known that $\angle B A C=60^{\circ}, A B$ $=2 A C$, point $P$ is inside $\triangle A B C$, and $P A=$ $\sqrt{3}, P B=5, P C=2$. Then the degree of $\angle A P C$ is $\qquad$ and the area of $\triangle A B C$ is . $\qquad$
4. $120^{\circ}, 3+\frac{7 \sqrt{3}}{2}$.

As shown in Figure 5, take the midpoint $D$ of $AB$, and the midpoint $F$ of $AP$. Rotate $\triangle APC$ clockwise by $60^{\circ}$ to $\triangle AED$, and connect $EF$, $DF$, and $PE$. Then $\triangle AEP$ is an equilateral triangle, $DE = PC = 2$, $EF = 1.5$, and $DF = \frac{BP}{2} = 2$.
In $\triangle DEF$, by the converse of the Pythagorean theorem, $\angle DEF = 90^{\circ}$.
Therefore, $\angle APC = \angle AED = 120^{\circ}$.
By $AD^{2} = (AF + DE)^{2} + EF^{2} = \left(\frac{\sqrt{3}}{2} + 2\right)^{2} + \left(\frac{3}{2}\right)^{2} = 7 + 2 \sqrt{3}$,

we know $S_{\triangle ABC} = \frac{1}{2} AB \cdot AC \sin 60^{\circ}$
$$
= \frac{\sqrt{3}}{2} AD^{2} = 3 + \frac{7 \sqrt{3}}{2} \text{. }
$$