Solution: Let point P be PC⊥AB, vertical foot be C, and let PC=x nautical miles.

In Rt△APC, ∫tan∠a = PC/AC, ∴ AC = PC/tan67.5 = 5x/ 12.

In Rt△PCB, ∫tan∠b = PC/BC, ∴ BC = x/tan36.9 = 4x/3.

∫AC+BC = ab = 2 1×5，∴5x/ 12+4x/3=2 1×5，x = 60。

∫sin∠b = PC/Pb, ∴ Pb = PC/sin ∠ b = 60/sin36.9 = 60× 5/3 =100 (nautical mile).

∴ The distance between City B and City P where the ship is located is 100 nautical mile.

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