A n-gon will be partinioned in the above manner. Let us number the partitions as 1, ..., i, i+1, ..., n/2. It can be seen that there are going to be n/2 such partitions for n-gon. As logically adding one more partition and two sides on top and bottom one can form another quadrilateral attached to the previous n-gon and in the process making a (n+3)-gon.

But there will always be two sides of the polygons acting as partitions on the extreme ends, i.e., the 1st and the (n/2)th partitions in this image. They are the non-diagonal partitions. So the number of diagonal partitions is (n/2-2).

So B is correct.