The "vector" part of this problem is to break down where the camel is walking into two components: a component which is only concerned with north-south travel, and a component which is only concerned with east-west travel.
You have the initial angle that the camel is walking: 15° west of due south; and it travels 23km. Using trigonometry, you can determine how much W of that 23km is just going westward, and how much S is just going southward. (Note that these two components don't add to 23km! Using Pythagoras' theorem, you should be able to see that W2 + S2 = (23km)2 .) So first, determine W and S. This will give you the position (call it C) of the camel after it's first blundering attempt to go to B.
Note that the usual convention is to describe west as being (negative east), and south as being (negative north). That is, because the camel is heading south and west, you should be able to describe its position in terms of a negative distance north, and separately a negative distance east.
The camel then starts going due north; it should be easy to find out where it ends up after that leg of the trip, because all of that leg is invested in north-south travel; you can just add that to the north-south component of the camel's position, to get its position D after the second leg.
Now you know where the camel is after the second leg of it's haphazard odyssey. Construct a right-angle triangle with the corners at B and D, and with the two shorter sides parallel to the x and y axes (that is, horizontal and vertical; or north-south and east-west). You should be able to apply Pythagoras' Theorem to find out how far away D is from B; that is, how far the camel has to travel to finally arrive where it was supposed to go. And for the direction, use trigonometry to find out the angle that it must travel, relative to east (which is the positive direction of the x axis).