So, I have square pyramid ABCDE, with E being the vertex. Segment ME is on lateral face AED and segment BP is on lateral face BEC. Segment EZ is perpendicular to the base ABCD of the pyramid. I'm trying to find the angle between skew segments BP and ME (the red and the yellow)
I assumed that the length of each edge is 1 for simplicity. Then MZ has length 1/2 since Z is the center of square ABCD (MZ would be half of CD). ME is not only a median but, since AED is an equilateral triangle, the altitude and angle bisector as well, so ME has length $\frac {\sqrt3} 2$. Then, the angle between ME and AD is $cos^{-1}(\frac {\sqrt3} 3)$.
And now I'm not sure where to go from here (or if I'm even approaching this correctly)
Your best bet is to setup an xyz-coordinate system as shown.
After determining the co-ordinates of each point in your figure (eg B = (0, 1, 0)), write down the equation of EM and BP.
The angle between them can then be found.
http://www.nabla.hr/PC-LinePlaneIn3DSp2.htm is a website that provides examples.