Collapsed partitions and ordinary partitions

Adopt the standard notation for integer partitions, writing $$\lambda_1^{a_1} \cdots \lambda_k^{a_k}$$ as shorthand for the partition $$a_1 \lambda_1 + \cdots + a_k \lambda_k$$ with parts $$\lambda_1 > \cdots > \lambda_k \geq 1$$ and multiplicities $$a_i \geq 1$$.

If $$\lambda = \lambda_1^{a_1} \cdots \lambda_k^{a_k} \vdash n$$, its collapsed image is the partition with multiple occurrences of parts removed, and is denoted by $$\underline{\lambda} := \lambda_1 \cdots \lambda_k$$. Introduce the set $$\underline{\mathcal{P}_n}=\{\,\underline{\lambda}: \,\lambda\vdash n\}$$ as well as the all-familiar $$\mathcal{P}_n=\{\,\lambda: \,\lambda\vdash n\}$$

QUESTION. Is this true? If so, is there a combinatorial proof? $$\#\underline{\mathcal{P}_n}=\sum_{j=0}^{n-1}\#\mathcal{P}_j.$$ CORRECTED. $$\sum_{\underline{\lambda}\in\underline{\mathcal{P}_n}}\text{length}(\underline{\lambda})=\sum_{j=0}^{n-1}\#\mathcal{P}_j.$$

Remark. For a related problem, see here:

The result does not seem to be stated correctly. For instance, $$\underline{\mathcal{P}_3}=\{ 3,21,1\}$$, yet $$p(0)+p(1)+p(2)=4$$, where $$p(n)=\#\mathcal{P}_n$$. It is true that $$\sum_{\lambda\in\underline{\mathcal{P_n}}}\ell(\lambda)=p(0)+p(1)+\cdots+ p(n-1),$$ where $$\ell(\lambda)$$ is the number of parts of $$\lambda$$. This is equivalent to the case $$k=1$$ of Enumerative Combinatorics, vol. 1, second ed., Exercise 1.80, which has a simple combinatorial proof.