= Solution
The standard <CW complex> structure on $\mathbb{CP}^n$ has one cell in every even dimension from $0$ to $2n$, and $\mathbb{CP}^k$ is its $2k$-skeleton. Collapsing that subcomplex leaves one zero-cell and one cell in each dimension $2i$ for $k+1\leq i\leq n$. All <cellular boundary formula>[cellular boundaries] vanish, so
$$
\boxed{H^q(\mathbb{CP}^n/\mathbb{CP}^k;\mathbb Z)\cong
\begin{cases}
\mathbb Z,&q=0\text{ or }q=2i,quad k+1\leq i\leq n,\\
0,&\text{otherwise}.
\end{cases}}
$$
Let $x\in H^2(\mathbb{CP}^n;\mathbb Z)$ be the usual generator. For $i>k$, choose $u_i$ whose pullback under the <quotient map> is $x^i$. Naturality of the <cup product> and the <cohomology ring of complex projective space> give
$$
u_i u_j=\begin{cases}
u_{i+j},&i+j\leq n,\\
0,&i+j>n.
\end{cases}
$$
Together with the unit, this determines the ring; equivalently, its reduced part is the ideal $(x^{k+1})/(x^{n+1})$ with the inherited multiplication. This is the <cohomology ring of a collapsed projective subspace>.
If $n=k+1$, the quotient has just a zero-cell and a $2n$-cell, so it is $S^{2n}$ and is a compact manifold. Conversely, suppose $n\geq k+2$ and the quotient is homotopy equivalent to a compact manifold. Its top cohomology is $\mathbb Z$, so that manifold must be closed, orientable, and $2n$-dimensional. But $H^{2n-2}\cong\mathbb Z$ while $H^2=0$, contradicting <Poincare duality>. Therefore
$$
\boxed{\mathbb{CP}^n/\mathbb{CP}^k\text{ is homotopy equivalent to a compact manifold exactly when }n=k+1.}
$$
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