Let $m \times a$ full-column matrix $B$, and define the projection matrix onto its column space as $P(B)=B(B'B)^{-1}B'$ which is $m\times m$ matrix.
Consider a rectangular matrix $A$, where $A$ is $m \times n$ and the rank of $A$ is $n$. More specifically, $A$ has a specific form such that $A=[A_1'|0']'$, where $A_1$ is $n\times n$ full-rank matrix and $0'$ denotes $n \times (m-n)$ zero matrix. Note that $a \leq n \leq m$ holds.
Define the fitted value as $P(B)A=B(B'B)^{-1}B'A$. In which case the fitted value becomes itself? In other words, I'm wondering about the conditions on $A$ or $B$ that make $P(B)A=B(B'B)^{-1}B'A=A$ hold.
Projection of any vector $\mathbf{x}$ that "lives" in the column space the design matrix $X$ will be $\mathbb{x}$ itself. Let us drop the intercept term for the sake of simplicity. As such, the $H$ matrix projects any vector into the space that is spanned by your independent variables $\mathbf{x}_1,...,\mathbf{x}_p$. Now, any vector that can be presented as a linear combination of $\mathbf{x}_1,...,\mathbf{x}_p$ belongs - by definition - to the space spanned by these vectors. The so called "fitted values" are nothing more but a linear combination of $\mathbf{x}_1,...,\mathbf{x}_p$ with the (unique vector of) coefficients $\hat{\beta}_1,..., \hat{\beta}_p$, hence the fitted values "live" in the column space of $X$. Hence, when you project the fitted vector you get the vector itself as the "distance" between the spanned space and the fitted values is zero.