How does the voltage divider in combination with a transistor cause clipping not to occur? Does it just set $V_b$ sufficiently high to be above the clipping point?

Basically, yes-- the voltage divider providing the DC bias at the input provides a DC offset sufficiently high that the most negative part of the signal will never go below a volt or so. If the DC bias is properly set, the transistor will always be ``on'', i.e., in the linear active region, for every part of the input signal's swing. In this case, there will be no clipping.