We have drawn k diagonals on a n-sided polygon.
Now, total edges = n(sides)+k(diagonals) = n+k
For each diagonal we draw, one extra region will be formed. Initially, there are 2 regions (entire region inside polygon and the external region)
total regions = k+2
Now, using the law, sum of degrees of all regions = 2*no of edges.
Here, r-1 regions will have degree of 4 (all the regions formed inside polygon) and the external region has degree (n+k) as it is bounded by (n) edges (sides of the polygon).
So, 4*(r-1)+(n)*1=2*e
4*(k+2-1)+(n)*1=2*(n+k)
4k+4+n=2n+2k
k=(n-4)/2
Option B is correct