Edit:

1) p^3 < p^5
(or) p < p^3

Satisfies only for +ve values of p. Also satisfies for -1 to 0.

InSufficient

2) p^2 < 1

-1<p<1 (p could take any value from -1 to +1)

Insufficient

Together only -1 to 0 is common.

C IMO

Missed out on one region the prev time,

p<p^3 is true when -1<p<0 or when p>1

Another way to get this is to factor statement 1:
p^3(1+p)(1-p)<0

The above statement is equal to zero when p=-1,0 and 1. Mark those points on a number line and think of the number line as being divided into four "zones": p<-1, -1<p<0, 0<p<1, p>1. You can then plug in test points from each of these zones to see if it satisfies the inequality. Or, just plug in something from p<-1, say p=-2. This would give you -8(-1)(3) which is positive, so every number less than -1 makes that expression positive. Each of the factors in the above expression changes signs when you cross its zero on the number line, so the signs of the zones would alternate too. + - + -, so the expression is negative and satisfies the inequality in the -1<p<0 and p>1 zones only. I hope this was clear. It's difficult to describe without drawing it out.

Anyway, statement 1 is therefore insufficient.

Statement 2) This is true when -1<p<1, so it is insufficient.

Both statements: -1<p<0 is the only region that satisfies both inequalities, so p must be negative,

so the answer is C