$P_n$ is given as $$P_n(f) = \sum_{\lambda \,\vdash\, n} (-1)^{\lambda_1} \prod \binom{\lambda_j}{\lambda_{j+1}} f_{j}^{\lambda_j}$$ where
- the sum is over partitions $\lambda = \lambda_1 \ge \lambda_2 \ge \cdots$ of $n$;
- $\lambda_{j+1} = 0$ if $j+1$ is greater than the number of parts in $\lambda$;
- I've adjusted the index from $f_{j+1}$ to $f_j$ because it appears that the increment was a hack to 1-index the variables in a 0-indexed programming language.
Then in the section Prototypic examples the text says
So let's define $P^k_n(f)$ as the $P$ transform of $f$ restricted to the partitions of $n$ with largest part $k$
Therefore $$P^k_n(f) = \sum_{\lambda' \,\vdash\, n-k} (-1)^{k} \binom{k}{\lambda'_1} f_1^k \prod \binom{\lambda'_j}{\lambda'_{j+1}} f_{j+1}^{\lambda_j}$$
A combinatorial interpretation of this can be obtained by defining a stalactite diagram (based on the English convention for Ferrers diagrams) to be a collection of cells where the top row is continuous and each cell in a lower row is directly below another cell. If we label the cells in the top row with $-f_1$ and the cells in row $k > 1$ with $f_k$ then $P_n^k(f)$ is the sum over all stalactite diagrams of $n$ cells in $k$ columns of the product of the cells in the diagram. This leads directly into the alternative formula $$P_n^k(f) = [x^n] \left(-f_1 x - f_1 f_2 x^2 - f_1 f_2 f_3 x^3 - \cdots \right)^k$$ and that answers the follow-up question as to the relationship with De Moivre polynomials: $A_{n,k}(a) = [x^n](a_1x + a_2x^2 + a_3x^3 + \cdots)^k$, so $$P_n^k(f) = A_{n,k}(-f_1, -f_1 f_2, -f_1 f_2 f_3, \ldots)$$