(1) ∠ NBP = 60, ∠ NAP = 30, so ∠ BPA = 30, so the isosceles triangle ABP, AB=BP.

AB=24*0.5= 12, so BP= 12 nautical mile.

(2) Because BC=0.5*24= 12, BC=BP= 12, and because ∠ CBP = 60 is a regular triangle CBP, CP= 12 nautical mile, so the distance between the ship and lighthouse P is/kloc-at this time.