For a positive integer n, let $$P_n$$ denote the product of the digits of n, and $$S_n$$ denote the sum of the digits of n. The number of integers between 10 and 1000 for which $$P_n$$ + $$S_n$$ = n is

Let n can be a 2 digit or a 3 digit number.

First let n be a 2 digit number.
So n = 10x + y and Pn = xy and Sn = x + y
Now, Pn + Sn = n
Therefore, xy + x + y = 10x + y , we have y = 9 .
Hence there are 9 numbers 19, 29,.. ,99, so 9 cases .  

Now if n is a 3 digit number.
Let n = 100x + 10y + z
So Pn = xyz and Sn = x + y + z
Now, for Pn + Sn = n ;  xyz + x + y + z = 100x + 10y + z ; so. xyz = 99x + 9y .
For above equation there is no value for which the above equation have an integer (single digit) value.

Hence option D.