Let \( f(N) = (3N^2 + 8N - 3)(pN - 1) \). We need to find the remainder when \( f(N) \) is divided by \( N + 1 \). According to the Remainder Theorem, the remainder of \( f(N) \) when divided by \( N + 1 \) is equal to \( f(-1) \).
To find \( f(-1) \), we start by calculating \( 3(-1)^2 + 8(-1) - 3 \):
\[ 3(-1)^2 + 8(-1) - 3 = 3(1) - 8 - 3 = 3 - 8 - 3 = -8 \]
Next, we calculate \( p(-1) - 1 \):
\[ p(-1) - 1 = -p - 1 \]
Now we can find \( f(-1) \):
\[ f(-1) = (-8)(-p - 1) = 8(p + 1) \]
According to the problem, the remainder \( f(-1) \) is given to be 24:
\[ 8(p + 1) = 24 \]
Dividing both sides by 8, we find:
\[ p + 1 = 3 \implies p = 2 \]
Now we want to find the remainder when \( p^4 \) is divided by 10. Calculating \( p^4 \):
\[ p^4 = 2^4 = 16 \]
Now we find the remainder of 16 when divided by 10:
\[ 16 \mod 10 = 6 \]
Thus, the remainder when \( p^4 \) is divided by 10 is:
\[ \boxed{6} \]