I have the following...
$\frac{\partial p}{\partial t}+\frac{\partial [p(1-p)]}{\partial x}=0$
$\Rightarrow \frac{\partial p}{\partial t}+(1-p)\frac{\partial p}{\partial x}-p\frac{\partial p}{\partial x}=0$
I am wondering if this is due to the chain rule? If not, why does it work this way?
It's not the chain rule, it's the Leibniz product rule for derivatives, which says that
$(f(x)g(x))' = f'(x) g(x) + f(x) g'(x), \tag 1$
where I use the abbreviated notation
$h'(x) = \dfrac{\partial h(x)}{\partial x}; \tag 2$
if we now set
$p(x) = f(x), \; 1 - p(x) = g(x), \tag 3$
then we see that
$(p(x)(1 - p(x))' = p'(x)(1 - p(x)) + p(x)(1 - p(x))' = p'(x)(1 - p(x)) - p(x)p'(x). \tag 4$