$M \space ? \space N$

Given that $\begin{aligned} N & = \frac{1}{4}+ \frac{3}{4} + \frac{5}{4} +\cdots + \frac{a} {4} \\& N = \frac{1}{4}\displaystyle\sum_{p=1}^{a} {(2p-1)} =\frac{ a^2}{4} \end{aligned}$ $\begin{aligned} M & = \frac{1}{8} + \frac{2}{8} +\frac{3}{8} +\cdots + \frac{a}{8} \\& M = \frac{1}{8} \displaystyle\sum_{p =1}^{a} p= \frac{1}{8} \frac{a(a+1)}{2} \\& M = \frac{a^2}{16} + \frac{a}{16} \implies M = \frac{1}{4} N + \frac{a}{16}\\& M = N -\frac{3N}{4} +\frac{a}{16} \implies M - N = \frac{-12N + a}{16} < 0 \end{aligned}$ Hence , $\boxed{N> M}$ since $-12N+a<0$

Given that

$\begin{cases} M = \displaystyle \sum_{k=1}^a \frac k8 \\ N = \displaystyle \sum_{k=1}^a \frac {2k-1}4 = 4\sum_{k=1}^a \frac k8 - 2\sum_{k=1}^a \frac 18 > 4\sum_{k=1}^a \frac k8 - 2\sum_{k=1}^a \frac k8 = 2M \end{cases}$

$\implies N > 2M \implies \boxed{N > M}$ .