Let's say we have
$$\sum_{k=1}^{p(N)}\prod_{j=1}^lf(i_{j_k},r_{j_k})$$
for some two-variable function $f(x,y)$.
Let $\lambda_k$ be the $k\text{th}$ partition of the integer $N$ into $l_k$ distinct parts.
$l_k$ is the number of distinct parts of the $k\text{th}$ partition of $N$.
$i_{m_k}$ is the number of times that the $m\text{th}$ distinct summand $r_{m_k}$ appears in $\lambda_k$ for some $1\le m\le l$.
So $\lambda_k=i_{1_k}r_{1_k}+\cdots+i_{l_k}r_{l_k}$.
$p(N)$ is the number of partitions of $N$
Now suppose instead of taking this summation over the partitions of some integer $N$, we take it over the partitions of every $N$.
Therefore we have
$$\sum_{N=1}^\infty\sum_{k=1}^{p(N)}\prod_{j=1}^lf(i_{j_k},r_{j_k})$$
Is there some combination of this sum taken over the partitions of all $N\ge 1$ that reduces to a series without partitions?
Let's recall the standard generating function case, where $f(i,r)=x^{ir}$ for a parameter $x$: in that case, $\prod_{j=1}^l f(i_{j_k},r_{j_k}) = x^{\sum_{j=1}^l i_{j_k}r_{j_k}} = x^N$, and so $$ \sum_{N=1}^\infty\sum_{k=1}^{p(N)}\prod_{j=1}^lf(i_{j_k},r_{j_k}) = \sum_{N=1}^\infty\sum_{k=1}^{p(N)} x^N = \sum_{N=1}^\infty p(N) x^N, $$ a generating function which is well-known: $$ \sum_{N=1}^\infty p(N) x^N = \prod_{k=1}^\infty (1 + x^k + x^{2k} + \cdots) = \prod_{k=1}^\infty \frac1{1-x^k}. $$
For a general function $f$, the same technique yields $$ \sum_{N=1}^\infty\sum_{k=1}^{p(N)}\prod_{j=1}^lf(i_{j_k},r_{j_k}) = \prod_{k=1}^\infty (1 + f(1,k) + f(2,k) + f(3,k) + \cdots) $$ (easiest to see by starting from the right-hand side and expanding it out); whether or not this simplifies further (or, indeed, even converges) depends upon the specific function $f$.