Don't you think it would help if you told us how {a,b,c,p} map to the bits in 0101? We are not mind readers.
As near as I can see, you should have two two-factor terms. You need to show us your work so that we can see where you are going wrong.
Do a sanity check. If only one row of the table is true, that means that there is exactly one combination of inputs that can make the result true. Looking at the original expression, that term has to be (a'bpc'), meaning that you have a=0 b=1 p=1 c=0 So now see if you can get the other term, (ab)'(b'c)', to be true from some combination of inputs other than the one above. Well, in order to make (xy)' a 1, we need to make (xy) a 0 which only takes making one or the other a 0. So we can trivially set a=0 and b=0 and not it doesn't matter what values b or p take on. Hence we have just identified FOUR rows for which the result is a 1. In fact, I believe you will find there are eight.