If is given that
$$
\begin{array}{r}
\frac{x^4}{(x-a)(x-b)(x-c)}=P(x)+\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c} \\
\Rightarrow x^4=(x-a)(x-b)(x-c) P(x)+A(x-b)(x-c) \\
+B(x-c)(x-a)+C((x-a)(x-b)
\end{array}
$$

At $x=0, a b c P(0)=b c A+c a B+a b C$


At $x=a, A=\frac{a^4}{(a-b)(a-c)}$, similarly
At $x=b, \quad B=\frac{b^4}{(b-c)(b-a)}$ and

So,
$$
\begin{aligned}
& P(0)=\frac{a^3}{(a-b)(a-c)}+\frac{b^3}{(b-c)(b-a)}+\frac{c^3}{(c-a)(c-b)} \\
& =\frac{a^3(c-b)+b^3(a-c)+c^3(b-a)}{(a-b)(b-c)(c-a)} \\
& =\frac{a^3(c-b)+b c\left(c^2-b^2\right)+a\left(b^3-c^3\right)}{(a-b)(b-c)(c-a)} \\
& =\frac{a^3+b c(c+b)-a\left(b^2+c^2+b c\right)}{(a-b(a-c)} \\
& =\frac{a\left(a^2-b^2\right)-c^2(a-b)-b c(a-b)}{(a-b)(a-c)} \\
& =\frac{\left(a^2+b a\right)-c^2-b c}{(a-c)}=a+b+c
\end{aligned}
$$

So, $P(0)+(a-b)(a-c) A=a+b+c+a^4$
Hence, option (d) is correct.