The answer is easier than it first appears.
We are provided with a correct balanced equation (I checked).
2 NABr + Ca(OH)2 = CaBr2 + 2 NaOH
This equation is telling us that we'll get 1 mole of CaBr2 for every 2 moles of NaBr, a molar ratio of 1/2. We'll only get half the moles that we put in of the NaBr. We have 14.72 moles of NaBr, and excess Ca(OH)2. So the calcium hydroxide is not a limiting reagent: all of the 14.72 moles of sodium bromide will react. The equation says we'll get 1/2 that number of moles of CaBr2. Therefore, we would expect 7.36 moles of CaBr2.
I hope that helps,