We have $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + cx + d$$

$$ \Rightarrow P'\left( x \right) = 4\,{x^3} + 3a{x^2} + 2bx + c$$

But $$P'\left( 0 \right) = 0 \Rightarrow c = 0$$

$$\therefore$$ $$P\left( x \right) = {x^4} + a{x^3} + b{x^2} + d$$

As given that $$P\left( { - 1} \right) < P\left( a \right)$$

$$ \Rightarrow 1 - a + b + d\,\, < \,\,1 + a + b + d \Rightarrow a > 0$$

Now $$P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx = x\left( {4{x^2} + 3ax + 2b} \right)$$

As $$P'\left( x \right) = 0,$$ there is only one solution $$x = 0,$$

therefore $$4{x^2} + 3ax + 2b = 0$$ should not have any real roots i.e. $$D < 0$$

$$ \Rightarrow 9{a^2} - 32b < 0$$

$$ \Rightarrow b > {{9{a^2}} \over {32}} > 0$$

Hence $$a,b > 0 \Rightarrow P'\left( x \right) = 4{x^3} + 3a{x^2} + 2bx > 0$$

$$\forall x > 0$$

$$\therefore$$ $$P(x)$$ is an increasing function on $$\left( {0,1} \right)$$

$$\therefore$$ $$P\left( 0 \right) < P\left( a \right)$$

Similarly we can prove $$P\left( x \right)$$ is decreasing on $$\left( { - 1,0} \right)$$

$$\therefore$$ $$P\left( { - 1} \right) > P\left( 0 \right)$$

So we can conclude that

Max $$P\left( x \right) = P\left( 1 \right)$$ and Min $$P\left( x \right) = P\left( 0 \right)$$

$$ \Rightarrow P\left( { - 1} \right)$$ is not minimum but $$P\left( 1 \right)$$ is the maximum of $$P.$$