How to calculate the equivalent weight of KMnO4?
Refer to the definition of equivalent (IUPAC Gold Book):

equivalent entity
Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.

In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom.
If hydrogen acts as a reducing or oxidizing agent, then either way $\pu{1 mol}$ of hydrogen atoms liberates or accepts $\pu{1 mol}$ of electrons:
$$
\begin{align}
\ce{0.5 H2 &amp;→ H+ + e-}\\
\ce{0.5 H2 + e- &amp;→ H-}
\end{align}
$$
That's why an equivalent of a redox agent is its amount which liberates or accepts $\pu{1 mol}$ of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.
Note that $\ce{KMnO4}$ participates in redox reaction
$$\ce{2K\overset{+7}{Mn}O4 + 16 H\overset{-1}{Cl} = 2 KCl + 2 \overset{+2}{Mn}Cl2 + 8 H2O + 5 \overset{0}{Cl}_2}$$
and the half-reaction for the reduction of manganese is
$$\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- → \overset{+2}{Mn}^2+ + 4H2O}$$
Also note that $n$-factor of permangante here is not the number of protons $\ce{H+}$, which are also used up in water formation; rather it's the number of transferred electrons, e.g. $n = 5$ and
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{5} = \pu{31.61 g mol-1}$$
Obviously, the equivalent mass of permanganate isn't a constant and depends on the $\mathrm{pH}$ of the reaction.
For example, in neutral medium half-reaction appears as
$$\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- → \overset{+4}{Mn}O2 + 4 OH-}$$
and there is no explicitly shown protons to count at all!
However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value:
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$