Why does the lone pair of phosphine not contribute to its dipole moment?
Let's see if we can use the following picture to help clarify things.



Let's look at the molecule $\ce{H3X-Z}$.  Let's say that the molecule is polarized such that the 2 electrons in the $\ce{X-Z}$ bond spend more time around $\ce{X}$ and less around $\ce{Z}$.  Let's also say that the $\ce{X-Z}$ bond is $\ce{sp^3}$ hybridized. Since the bond is polarized it will contribute to the molecular dipole moment.  Are you with me so far?

If we stretch that bond and look at the bond right before it breaks, it will still be polarized, but with a larger bond dipole moment because the electrons are still predominately located around $\ce{X}$ and the distance between those electrons and the electron depleted $\ce{Z}$ nucleus has increased.  If we stretch the bond a bit further it will break and we will wind up with two electrons around $\ce{X}$.  These electrons around $\ce{X}$ are still in an $\ce{sp^3}$ orbital as shown in the bottom part of the picture.  The orbital and the electrons in it still "point" towards where $\ce{Z}$ used to be.  The orbital and the electrons in it still have directionality.  They still contribute to a dipole moment.  Can you see this?

Let's continue assuming you can.  Let's repeat all that we've done above with one change.  Let's now say that the $\ce{X-Z}$ bond is $\ce{s}$ hybridized.  Now when we get to the last step and break the $\ce{X-Z}$ bond, the electrons remain in an $\ce{s}$ orbital.  An $\ce{s}$ orbital is spherically symmetric.  It has no directionality, it doesn't "point" in any specific direction because it is spherically symmetric.  So in this case we do not have a contribution to the molecule's dipole moment from the electrons in the spherically symmetric orbital.