I
I R K
+ S I R
-------
S K I P
Yes, S has to be 1, because the sum of two 3-digit numbers and a 1-digit number can't be more than 1999.
I
I R K
+ 1 I R
-------
1 K I P
Now with S=1, K (in SKIP) has to be either 1 or 0; but it can't be 1, because S is 1; so K is 0.
I
I R 0
+ 1 I R
-------
1 0 I P
I'll show you a path to finishing the problem from here and let you do the actual work....
In the tens column, with R plus I giving digit I in the sum, there are only two possible values for R; but one of them has already been used. So you know what R has to be.
Now in the hundreds column I plus S (=1) giving K (=0) in the sum leaves only two possible values for K; but one of them has already been used. So now you know what K has to be.
Then the only thing left to do is perform the addition to find the value of P.