We need the fact that the sum of powers, $j\ge0$,
$$
S(j)=\sum_{x=0}^{p-1}x^j
$$
is $\equiv 0\pmod{p}$, unless $j$ is positive and divisible by $(p-1)$, and in that latter case we have
$S(j)\equiv -1\pmod{p}$. This follows from the fact that when $j>0$, the non-zero terms
can be reordered (according to ascending powers of a primitive root) to form a geometric sum of $p-1$ terms. Modulo $p$ such a sum is either all ones, or the sum of a full set of roots of unity of order dividing some integer $d\mid p-1$. In the latter case the sum is congruent to zero modulo $p$.
Another fact that we need is the congruence (Euler's formula)
$$
\rho(x)\equiv x^{(p-1)/2}\pmod p.
$$
We need to compute the residue class modulo $p$ of the character sum
$$
S=\sum_{x=0}^{p-1}\rho(1-x^4)\equiv\sum_{x=0}^{p-1}(1-x^4)^{\frac{p-1}2}.
$$
Here $(p-1)/2=2m$. By the binomial formula
$$
(1-x^4)^{2m}=\sum_{j=0}^{2m}(-1)^j{2m\choose j}x^{4j}.
$$
Summing over $x$ we get
$$
S\equiv\sum_{j=0}^{2m}(-1)^j{2m\choose j}S(4j).
$$
Here $S(4j)$ is non-zero modulo $p$ only, when $j= m$ or $j=2m$.
Putting these pieces together we get
$$
S\equiv(-1)\left((-1)^m{2m\choose m}+1\right).
$$
The claim follows from this and yet another part of this exercise stating that
$$
N(y^2+x^4=1)=p-1+2a.
$$
The claim of that part of the exercise is derived in the same way as the other relations between Jacobi sums and the numbers $N(x^n+y^m=1)$ in this chapter, namely
$$
N(y^2+x^4=1)=\sum_{i=0}^3\sum_{j=0}^1 J(\chi^i,\rho^j).
$$
