1
It means $j_1, \ldots, j_d$ are nonnegative and sum to $p$.
and $$\binom{p}{j_1, \ldots, j_d} = \frac{p!}{j_1! j_2! \ldots j_d!}=\frac{p!}{\prod_{i=1}^d j_i!}$$
The multinomial coefficient.
Note that $$(u^Tv)^p = \left(\sum_{i=1}^d u^{(i)}v^{(i)} \right)^p$$
