Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$ such that $x = 0$ is the only real root of $P'(x) = 0$. If $P(-1) < P(1)$, then in the interval $[-1, 1]$.

Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$

$P'(x) = 4x^3 + 3ax^2 + 2bx + c$

Since $x = 0$ is the only real root of $P'(x) = 0$, we have $P'(0) = 0 \Rightarrow c = 0$.

$P'(x) = x(4x^2 + 3ax + 2b)$

For $x = 0$ to be the only real root, the quadratic equation $4x^2 + 3ax + 2b = 0$ must have no real roots (or $x=0$ is a repeated root, which implies $a=0, b=0$). In either case, $4x^2 + 3ax + 2b > 0$ for all $x \neq 0$ because the leading coefficient is positive.

Thus, $P'(x) < 0$ for $x < 0$ and $P'(x) > 0$ for $x > 0$.

This means $P(x)$ is strictly decreasing on $[-1, 0]$ and strictly increasing on $[0, 1]$.

The minimum value of $P(x)$ in $[-1, 1]$ occurs at $x = 0$. Therefore, $P(-1)$ is not the minimum.

The maximum value of $P(x)$ in $[-1, 1]$ must occur at one of the endpoints, $x = -1$ or $x = 1$.

Since it is given that $P(-1) < P(1)$, the maximum value is $P(1)$.

Hence, $P(-1)$ is not the minimum but $P(1)$ is the maximum of $P$.

Answer: $P(-1)$ is not minimum but $P(1)$ is the maximum of $P$