using P4(g) + 6Cl2(g) → 4PCl3(g) determine what bonds are broken and what bonds are formed.?

The equation with structure formulae of the reactants and products are shown below.

According to the equation, 6 moles of P-P bonds and 6 moles of Cl-Cl bonds are broken, and 4 × 3 moles of P-Cl bonds are formed.
Energy absorbed in bond breaking = (201 × 6 + 242 × 6) kJ = 2658 kJ
Energy released in bond formation = E(P-Cl) × 4 × 3 = 12 E(P-Cl)

ΔrH = (Energy absorbed in bond breaking) - (Energy released in bond formation)
-1206.9 kJ = (2658 kJ) - 12 E(P-Cl)
Bond dissociation enthalpy for the P-Cl bond, E = (2658 + 1206.9)/12 kJ = 322 kJ

Bond energies.....

P4(g) + 6Cl2(g) --> 4PCl3(g)

It's tempting to say that there are 4 P-P bonds in P4. But that isn't correct. P4 is tetrahedral. Draw a tetrahedron. There are 4 sides, 4 vertices and 6 edges. Each edge is a bond between P atoms.

Breaking.... 6 P-P bonds and 6 Cl-Cl bonds
Making ...... 12 P-Cl bonds

Bond breaking is endothermic (ΔH is positive) and bond making is exothermic (ΔH is negative).

ΔH(reaction) = ΣBE
ΔH(reaction) = 6(P-P) + 6(Cl-Cl) + 12(P-Cl)
-1206.9 = 6(201) + 6(242) + 12(-x)
=== some algebra goes here ===
x = 322
The bond energy for P-Cl is 322 kJ/mol.

One source lists the P-Cl bond energy as 326 kJ/mol:
http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html