There is no need to calculate the discriminant to find this Galois group if $N \geq 2$. Let $f(x) = x^5 - Npx + p$ with $N \geq 2$ and let $G$ be the Galois group of $f(x)$. We have
\begin{equation*}
\begin{aligned}
f(-Np) &= -N^5p^5 + N^2p^2 + p < 0\\
f(0) &= p > 0 \\
f(1) &= 1 + p - Np < 0.
\end{aligned}
\end{equation*}
Since $f(x)$ has two sign changes, it must have at least two real roots. Since the derivative $f'(x) = 5x^4 - Np$ has only two real roots, it follows that $f(x)$ cannot have all real roots. But the nonreal roots come in conjugate pairs, and so $f(x)$ must have three real roots and two nonreal roots. Therefore the transposition which interchanges the two nonreal roots is in $G$, hence $G$ is isomorphic to $S_5$.
If $N = 1$ then it depends on how large $p$ is. You can show that if $p \geq 13$ then there are three real roots, and then again $G$ is isomorphic to $S_5$. Now if $p < 13$, you can check by graphing that there is only one real root and then $G$ contains a $5$-cycle and a double transposition and is thus isomorphic to either $A_5$ or $D_5$.
Returning to the question about the discriminant, let $f(x) = x^5 + ax + b$ be an irreducible quintic. Let $\alpha_1,\ldots,\alpha_5$ be the complex roots of $f(x)$. Then $f(x) = (x - \alpha_1)\cdots(x - \alpha_5)$, and so taking the derivative with the product rule gives
$$f'(x) = \sum_{i=1}^{5}\prod_{\substack{1 \leq j \leq 5\\ i\neq j}}(x - \alpha_j),$$
hence
$$f'(\alpha_i) = \prod_{\substack{1 \leq i,j \leq 5\\ i\neq j}}(\alpha_i - \alpha_j)$$
You can check that $\Delta = \prod_{i=1}^{5}f'(\alpha_i)$.
We also have $f'(x) = 5x^4 + a$, hence $f'(\alpha_i) = 5\alpha_i^4 + a$. Therefore
$$\Delta = \prod_{i=1}^{5}(5\alpha_i^4 + a).$$
Expanding the above product out gives a formula for $\Delta$ in terms of the elementary symmetric polynomials in $\alpha_1^4,\ldots,\alpha_5^4$. Reexpressing these in terms of the elementary symmetric polynomials in $\alpha_1,\ldots,\alpha_5$ thus gives a formula for $\Delta$ in terms of the coefficients of $f(x)$.
