Step 1: Think about this in terms of eigenvalues.
If $P^3 = I$, every eigenvalue $\mu$ of $P$ must satisfy $\mu^3 = 1$. The numbers satisfying this are the cube roots of unity: $\mu = 1$, $\mu = \omega$, $\mu = \omega^2$, where $\omega = e^{2\pi i/3}$. Any relation that always holds for the matrix must also hold when we plug in each of these scalar values.
Step 2: Test option (A), $P^5=P$, using $\mu=\omega$.
$\mu^5 = \omega^5 = \omega^{3+2} = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$.
For the option to hold this should equal $\mu=\omega$, but $\omega^2 \neq \omega$. So option (A) fails.
Step 3: Test option (B), $P^5=P^{-1}$, using $\mu=\omega$.
We already found $\mu^5 = \omega^2$.
Also $\mu^{-1} = \omega^{-1}$, and since $\omega^3=1$ we have $\omega \cdot \omega^2 = \omega^3 = 1$, so $\omega^{-1} = \omega^2$.
So $\mu^5 = \omega^2 = \mu^{-1}$. This holds. Checking $\mu=1$ separately: $1^5=1=1^{-1}$, which also holds. So option (B) is consistent for every eigenvalue.
Step 4: Test options (C) and (D) the same way.
Option (C) $P^4=P^{-1}$: $\mu^4 = \omega^{3+1}=\omega$, but $\mu^{-1}=\omega^2 \neq \omega$. Fails.
Option (D) $P^4=P^2$: $\mu^4=\omega$, but $\mu^2=\omega^2 \neq \omega$. Fails.
Step 5: Conclusion.
Only option (B), $P^5=P^{-1}$, is consistent for every eigenvalue satisfying $\mu^3=1$, so it is the relation that must hold for the matrix $P$ itself.
Final Answer:
\[ \boxed{P^5 = P^{-1}} \]