n can be a 2 digit or a 3 digit number.
Case (1) Let n be a 2 digit number.
Let n = 10x + y
$P_{n}$ = xy and $S_{n}$ = x + y
Now, $P_{n}$ + $S_{n}$ = n
xy + x + y = 10x + y
xy = 9x = 9
There are 9 two digit numbers (19, 29, 29, … ,99) for which y = 9
Case (2) Let n be a 3 digit number.
Let n = 100x + 10y + z
$P_{n}$ = xyz and $S_{n}$ = x + y + z
Now, $P_{n}$ +$S_{n}$ = n xyz + x + y + z = 100x + 10y + z
xyz = 99x + 9y
z = $\frac{99}{y} + \frac{9}{x}$
From the above expression, 0 < x, y < 9
But, we cannot find any value of x and y, for which z is a single digit number.
There are no 3 digit numbers which satisfy$P_{n}$ +$S_{n}$ = n
Hence, option (4)9 is the Answer.