这道题怎么解---```text
Question Stem:
如图，在三棱锥P-ABC的平面展开图中，CD∥AB，AB⊥AC，AB=2AC=2，CD=√13，cos∠BCF = 4√65/65，则三棱锥P-ABC外接球表面积为_______。
Mathematical Formulas/Equations and Conditions:
CD∥AB
AB⊥AC
AB = 2AC = 2
CD = √13
cos∠BCF = 4√65/65
Geometric Figure Description:
Type: Plane net of a tetrahedron P-ABC.
Main Elements:
Points labeled: A, B, C, D(P), E(P), F(P).
Line segments connecting these points. A is connected to B, C, D(P), E(P). B is connected to A, C, E(P), F(P). C is connected to A, B, D(P), F(P). D(P) is connected to A, C. E(P) is connected to A, B. F(P) is connected to B, C.
The figure shows triangle ABC as a base, with triangles ACD(P), ABE(P), BCF(P) attached to its sides. The points D(P), E(P), F(P) represent the vertex P when the faces PAC, PAB, PBC are unfolded onto the plane.
In this net, AD(P) = AP, AE(P) = AP, CD(P) = CP, CF(P) = CP, BE(P) = BP, BF(P) = BP.
The conditions relate features in the net to properties of the original tetrahedron. CD is the segment CD(P) in the net. ∠BCF is the angle ∠BCF(P) in the net.
```