= Solution
Here projection means <Hermitian> <orthogonal projection>: $P=P^*=P^2$ with $\operatorname{tr}P=k$, as indicated by the unitary conjugation in the question. These matrices represent the <complex Grassmannian as orthogonal projections>, a compact manifold of complex dimension $k(n-k)$. The constants $c_i$ are real. Let $A^*=-A$ and use the given tangent directions $[A,P]$. Cyclicity of the <matrix trace> gives
$$
df_P([A,P])=\operatorname{tr}(C[A,P])=\operatorname{tr}([P,C]A).
$$
The <commutator> $B=[P,C]$ is skew-Hermitian. If this derivative vanishes for every skew-Hermitian $A$, choose $A=B$ to obtain
$$
0=\operatorname{tr}(B^2)=-\operatorname{tr}(B^*B),
$$
so $B=0$. Conversely $[P,C]=0$ makes every tangent derivative zero. Because the diagonal entries of $C$ are distinct, a commuting <matrix> $P$ is diagonal. Its idempotent entries are zero or one. Thus the critical projections are exactly
$$
\boxed{P_I=\operatorname{diag}(\mathbf1_{1\in I},\ldots,\mathbf1_{n\in I}),\qquad |I|=k,}
$$
with $\binom nk$ <critical points> and values $f(P_I)=\sum_{i\in I}c_i$.
To compute the <Hessian matrix>, write $E=\operatorname{span}\{e_i:i\in I\}$ and $E^\perp=\operatorname{span}\{e_j:j\notin I\}$. A neighboring plane is the graph of a complex <linear map> $Z:E\to E^\perp$. With $W=\binom I Z$, its orthogonal projection is $P(Z)=W(I+Z^*Z)^{-1}W^*$. Writing $C_E,C_{E^\perp}$ for the diagonal blocks, we get
$$
f(P(Z))=\operatorname{tr}\bigl(C_E(I+Z^*Z)^{-1}\bigr)+\operatorname{tr}\bigl(C_{E^\perp}Z(I+Z^*Z)^{-1}Z^*\bigr).
$$
Expanding $(I+Z^*Z)^{-1}=I-Z^*Z+O(\|Z\|^4)$ yields
$$
f(P(Z))=f(P_I)+\sum_{i\in I,\ j\notin I}(c_j-c_i)|z_{ji}|^2+O(\|Z\|^4).
$$
Each complex coordinate has two real components, each with Hessian coefficient $2(c_j-c_i)$. Every coefficient is nonzero, proving that $f$ is a <Morse function>. Its negative directions correspond to $j<i$. For $I=\{i_1<\cdots<i_k\}$ there are $i_a-a$ complementary indices below $i_a$, giving
$$
\boxed{\operatorname{ind}(P_I)=2\#\{(i,j):i\in I,\ j\notin I,\ j<i\}=2\sum_{a=1}^k(i_a-a).}
$$
This is the <diagonal trace Morse function on a complex Grassmannian>. Its critical values need not be distinct; ties between sums do not make its critical points degenerate.
All <Morse indices> are even. Choose a <Morse-Smale gradient flow> and form its integral <Morse chain complex>. The odd-degree chain groups vanish, so every differential vanishes. Equivalently the <Morse handle-attachment theorem> supplies only even-dimensional cells. The <integral homology of complex Grassmannians> is therefore
$$
\boxed{H_{2r}(\operatorname{Gr}(k,n);\mathbb Z)=\mathbb Z^{N_r},\quad H_{2r+1}(\operatorname{Gr}(k,n);\mathbb Z)=0,}
$$
where $N_r$ counts the subsets with $\sum_a(i_a-a)=r$, and $0\leq r\leq k(n-k)$. In particular there is no torsion. The nondecreasing sequence $i_a-a$ lies between zero and $n-k$, so reversing it identifies these subsets with <partitions of an integer> $r$ fitting in a $k$ by $(n-k)$ rectangle. The <Poincaré polynomial> is explicitly
$$
\sum_{|I|=k}t^{2\sum_a(i_a-a)}.
$$
For the permitted example $\operatorname{Gr}(2,4)$, the subsets $12,13,14,23,24,34$ have indices $0,2,4,4,6,8$. Thus
$$
\boxed{H_0=H_2=H_6=H_8=\mathbb Z,\quad H_4=\mathbb Z^2,\quad H_{\mathrm{odd}}=0.}
$$
The general calculation above applies throughout $2\leq k\leq n-2$; it does not reduce to any of the excluded rank-one cases.
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