We have to find the number of solutions to the equation:- $$x^e \equiv c \mod p$$ where $c \not\equiv 0\mod p$. For $c=1$, we can prove that the above has $\gcd(e,p-1)=d$ solutions in the following way:-
Assume that $g$ is a primitive root of $p$. Hence there exists $0<k \le p-1$ such that $$g^k \equiv x\mod p \implies g^{ke} \equiv 1\mod p \implies (p-1)|ke$$ Let $d=\gcd(e,p-1)$. Therefore $\left(\frac{p-1}d\right) \mid k \cdot\left(\frac ed \right)$. Now, we know that $\gcd \left(\frac{p-1}d, \frac ed \right)=1$. Therefore $\frac{p-1}d$ must divide $k$. Therefore $k \in \left\{\frac{p-1}d, 2\left(\frac{p-1}d\right), \cdots,d\left(\frac{p-1}d\right)\right\}$, (since $0<k\le p-1$). Hence for $c=1$, there are $d$ solutions.
Now, I am asked to show that if a solution to $x^e \equiv c\mod p$ exists, then there must be $d$ unique solutions to the equation. I have shown that a solution exists iff $c=g^{dn}\mod p$, where $d=\gcd(e,p-1)$ and $n \in \{1,2,\cdots,p-1\}$. For this general equation, we must solve the follwing equation and show that it has $d$ unique solutions:-
$$ke \equiv nd \mod p-1 \implies k\left(\frac ed\right) \equiv n \mod \frac{p-1}d$$ How do I show that this has exactly $d$ unique solutions?
Suppose that $$ x^e \equiv c \pmod p $$ has a solution, say $x \equiv a$. If $b$ is any solution, you can find some $r$ such that $ar \equiv b$. Then it follows that $$ (ar)^e \equiv c \equiv a^e \pmod p. $$ Canceling $a^e$, we find that $$ r^e \equiv 1 \pmod p. $$ So $r$ must be one of the $d$ solutions to $x^e \equiv 1$.
Conversely, you can see that if $r^e \equiv 1$, then $(ar)^e \equiv c$. Hence the solutions to $x^e \equiv c$ are the values $ar$ for each of the $d$ possible values of $r$.