On one processor we have $$T_1 = (1-f)W + fW = W$$ On $p$ processors we have $$T_p = (1-f)W + \frac{fW}{p},$$ resulting in a speedup of $$\frac{T_1}{T_p} = \frac{W}{(1-f)W+fW/p}$$

As p goes to infinity, $fW/p$ goes to zero, and the maximum speedup is $$\frac{1}{1-f},$$ meaning that if if $f = 0.99$ (all but $1\%$ parallelizable), the maximum speedup is $1/(1-.99)=100$!