= Solution
The <Grassmann field> $c$ is odd and the <adjoint scalar field> $\phi$ is even. The two printed signs are consistent with a <right-acting BRST differential>, whose <graded Leibniz rule> is
$$
Q(XY)=X(QY)+(-1)^{|Y|}(QX)Y.
$$
The bracket between two odd fields is graded, so $[c,c]=2c^2$. Applying this rule to the ordinary <Lie bracket> $[c,\phi]=c\phi-\phi c$ gives
$$
Q^2\phi=[Qc,\phi]+[c,Q\phi].
$$
The second bracket here is graded because both its entries are odd. The <graded Jacobi identity> gives $[c,[c,\phi]]=\tfrac12[[c,c],\phi]$. Therefore
$$
\boxed{Q^2\phi=-\frac12[[c,c],\phi]+\frac12[[c,c],\phi]=0.}
$$
\b[The side of the odd derivation is essential.] With the usual left <graded Leibniz rule> and the same two printed signs, the result would be $-[[c,c],\phi]$, which is generally nonzero. A left-acting convention must reverse one of those signs. All subsequent <BRST symmetry> formulas here use the right-acting convention.
Choose an anti-Hermitian basis $T^a$ with $[T^b,T^c]=f^{abc}T^a$, an invariant positive <invariant bilinear form on a Lie algebra>, and the <adjoint covariant derivative> $D_\mu X=\partial_\mu X+[A_\mu,X]$. Couplings are absorbed into this convention; restoring $D_\mu=\partial_\mu+g[A_\mu,\cdot]$ multiplies each ghost <interaction vertex> below by $g$. The associated <BRST charge> acts as $QA_\mu=-D_\mu c$, $Q\bar c=ih$, and $Qh=0$.
Write $F^a=\partial\cdot A^a+(n\cdot\partial\phi^a)/\sqrt2$. For the <gauge-fixing fermion> $\Psi=\int\bar c^a(F^a-ih^a/2)$, the right <graded Leibniz rule> gives
$$
Q\Psi=\int d^4x\left[\frac12h^ah^a+ih^aF^a+\bar c^a(QF)^a\right],\qquad
QF=-\partial^\mu D_\mu c-\frac1{\sqrt2}n\cdot\partial[\phi,c].
$$
The <Gaussian functional integral> over the <Nakanishi-Lautrup field> produces the positive <gauge fixing> term $F^aF^a/2$. The ghost operator is $-\partial\cdot D-(n\cdot\partial)\operatorname{ad}_\phi/\sqrt2$.
Now use the canonical free <kinetic terms> $F_{\mu\nu}^aF_{\mu\nu}^a/4+(\partial\phi^a)^2/2$. At nonzero momentum in <Euclidean space> $p$, put $\eta=n\cdot p$. The quadratic kernel for $(A_\mu^a,\phi^a)$, per color, is
$$
K(p)=\begin{pmatrix}
p^2\delta_{\mu\nu}&\eta p_\mu/\sqrt2\\
\eta p_\nu/\sqrt2&p^2+\eta^2/2
\end{pmatrix}.
$$
The transverse gauge-field kernel plus the gauge-fixing longitudinal term has become $p^2\delta_{\mu\nu}$. The scalar <quantum field theory propagator> is the inverse <Schur complement>, not merely the inverse of the scalar diagonal entry:
$$
K_{\phi\phi}-K_{\phi A}K_{AA}^{-1}K_{A\phi}
=p^2+\frac{\eta^2}{2}-\frac{\eta^2 p^2}{2p^2}=p^2.
$$
Hence the <free adjoint-scalar propagator in scalar-dependent gauge fixing> is
$$
\boxed{\langle\phi^a(p)\phi^b(-p)\rangle_0=\frac{\delta^{ab}}{p^2}.}
$$
For completeness the mixed <quantum field theory propagator> is $\langle A_\mu^a\phi^b\rangle_0=-\delta^{ab}\eta p_\mu/(\sqrt2 p^4)$; ignoring this mixing would give an incorrect scalar answer.
\b[The scalar propagator is independent of the gauge vector $n$, not of the momentum component parallel to $n$.] For a unit $n$, $p^2=p_\perp^2+p_\parallel^2$, and the free scalar <quantum field theory propagator> still depends on $p_\parallel$. Thus the literal momentum-independence clause in the PDF is false for the standard minimally coupled massless scalar action; the cancellation above establishes the natural gauge-vector-independence statement. The usual massless <zero mode in field theory> at $p=0$ needs a separate infrared prescription.
Finally, <integration by parts> gives the ghost action in an unambiguous convention:
$$
S_{\mathrm{gh}}=\int d^4x\left[(\partial_\mu\bar c^a)(\partial_\mu c^a)+f^{abc}(\partial_\mu\bar c^a)A_\mu^b c^c+\frac{f^{abc}}{\sqrt2}(n\cdot\partial\bar c^a)\phi^b c^c\right].
$$
Use $e^{ip\cdot x}$ for the <Fourier transform> of every field, with all momenta incoming. Let the antighost carry color $a$ and momentum $q$, the boson color $b$ and momentum $k$, and the ghost color $c$ and momentum $p$, so $q+k+p=0$. Expansion of $e^{-S_{\mathrm{gh}}}$ gives the <ghost vertices in scalar-dependent gauge fixing>
$$
\boxed{\bar c^a(q)A_\mu^b(k)c^c(p):\ -i f^{abc}q_\mu,\qquad
\bar c^a(q)\phi^b(k)c^c(p):\ -\frac{i}{\sqrt2}f^{abc}(n\cdot q).}
$$
The free <Faddeev-Popov ghost field> propagator is $\delta^{ab}/p^2$ and every closed <ghost loop> contributes a minus sign. There are no further ghost <interaction vertices> in this gauge. Factors of $i$ and an overall ghost-vertex sign depend on the <Fourier transform> and ghost-ordering conventions; the displayed ghost action fixes both here.
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