Is the friction on a parallelepiped with mass m equal to the friction on a dice with the same mass? Because the overhang of the parallelepiped causes an irregular distribution of the normal force on the ground.
As outlined by OP in a comment, we are dealing with an object lying on a surface with one side fully in contact with that surface.
Kinetic friction doesn't depend on area. In only depends on normal force (the pressure on the surface):*
$\mu_k$ is the friction coefficient and can be thought of as a constant (depends on the two materials, on their interaction, roughnesses etc.).
Two objects with equal masses have the same weight. When lying freely on a surface (no extra pushes or burdens), that weight equals the normal force (due to Newton's 1st law).
So, it doesn't matter if the object is a parallelepipedum, a cube, a box or other - meaning if the contacting surface is triangular, square, rectangular or other. If their masses are the same, then the normal force is the same and then kinetic friction is the same.**
* This kinetic-friction model applies for not-too-large normal forces per area (called normal pressure), so it is useful in many daily-life scenarios, and I'm assuming also in yours. If you are dealing with, say, cutting tools in machining or turning or similar, then we shouldn't use this model.
** When saying this, I am referring to the total kinetic friction. While the total kinetic frictions are equal for any objects of the same mass, regardless of contact area, the distributions of the frictions may not be equal. More weight in one side of an object (like in the parallelepipedum you mention) will cause a larger normal force in this side and thus, according to the friction model above, a larger kinetic friction on this area, whereas an object with symmetric mass (like a cube) will give a different and symmetric kinetic-friction distribution.