Solution:
We start with a simple fact:

Lemma: If $b$ is an integer not divisible by $p$ then there is an integer $s$ so that $s b$ has the remainder $l$ when divided by $p$.
For a proof, just note that numbers $b, 2b, \ldots, (p-1)b$ have distinct non-zero remainders when divided by $p$, and hence one of them is equal to $1$.

We prove that if $x, y = 0, 1, 2, \ldots, p-1$ and $p$ divides $x^{5} - y^{5}$, then $x = y$.
Indeed, assume that $x \neq y$. If $x = 0$, then $p \mid y^{5}$ and so $y = 0$, a contradiction.
To this point we have $x, y \neq 0$. Since
$$p \mid (x-y)\left(x^{4} + x^{3}y + x^{2}y^{2} + x y^{3} + y^{4}\right)$$
and $p \nmid (x-y)$,
we have
$$\begin{aligned} & p \mid \left(x^{2} + y^{2}\right)^{2} + x y \left(x^{2} + y^{2}\right) - x^{2} y^{2}, \text{ and so } \\ & p \mid \left(2\left(x^{2} + y^{2}\right) + x y\right)^{2} - 5 x^{2} y^{2} \end{aligned}$$
As $p \nmid x y$, from the lemma we find an integer $s$ so that $s x y = k p + 1$, $k \in \mathbb{N}$. Then
$$p \mid \left[s\left(2 x^{2} + 2 y^{2} + x y\right)\right]^{2} - 5\left(k^{2} p^{2} + 2 k p + 1\right)$$
and so $p \mid z^{2} - 5$, where $z = s\left(2 x^{2} + 2 y^{2} + x y\right)$, a contradiction.
Consequently $x = y$.

Since we have proved that numbers $0^{5}, 1^{5}, \ldots, (p-1)^{5}$ have distinct remainders when divided by $p$, the same goes for the numbers $0^{5} + a, 1^{5} + a, \ldots, (p-1)^{5} + a$ and the conclusion can be reached easily.