= Solution
Both the <D meson> and <kaon> are spin-zero <pseudoscalars>. The <strong interaction> states obey <parity symmetry in quantum field theory>. Consequently an <axial current> <matrix element> between them would have to be a <pseudovector> formed from only $p$ and $k$, which is impossible: an <totally antisymmetric tensor> would require more <linearly independent vectors>. Thus
$$
\langle K^+(k)|\bar c\gamma^\mu\gamma^5s|\bar D^0(p)\rangle=0.
$$
<Lorentz covariance> then leaves two <linearly independent vectors> for the <vector current> <matrix element>, giving the <pseudoscalar-to-pseudoscalar form factors>
$$
\langle K^+(k)|\bar c\gamma^\mu s|\bar D^0(p)\rangle
=(p+k)^\mu f_+(q^2)+(p-k)^\mu f_-(q^2),\qquad q=p-k.
$$
The vanishing <axial current> here follows from <strong interaction> <parity symmetry in quantum field theory>, not <parity symmetry in quantum field theory> of the <weak interaction>.
To obtain the requested decay formula, use <naive factorization of a nonleptonic meson decay>: approximate the four-quark <matrix element> by the product of the $\bar D^0\to K^+$ current <matrix element> and the vacuum-to-<pion> <matrix element>. This is an additional hadronic approximation; tree-level weak vertices alone do not establish it, and nonfactorizable <Quantum chromodynamics> effects can change the result. The vacuum-to-<pion> <vector current> <matrix element> vanishes by <parity symmetry in quantum field theory>, while the specified <pion decay constant> normalization gives
$$
\langle\pi^-(q)|\bar d\gamma_\mu(1-\gamma^5)u|0\rangle=+i\sqrt2 F_\pi q_\mu.
$$
The $\sqrt2$ cancels the $1/\sqrt2$ in the <four-fermion interaction>. Up to an irrelevant overall phase, the <scattering amplitude> is therefore
$$
\mathcal M=G_F V_{cs}V_{ud}^*F_\pi\left[(m_D^2-m_K^2)f_+(q^2)+q^2f_-(q^2)\right].
$$
For $m_\pi=0$, $q^2=0$, so only $f_+(0)$ survives. Integrating the <two-body decay phase space> in the <D meson> rest frame gives
$$
|\boldsymbol k|=\frac{m_D^2-m_K^2}{2m_D},\qquad
\Gamma=\frac{|\boldsymbol k|}{8\pi m_D^2}|\mathcal M|^2.
$$
Hence
$$
\boxed{A=\frac{(m_D^2-m_K^2)^3}{16\pi m_D^3},\qquad
\Gamma=\frac{G_F^2(m_D^2-m_K^2)^3}{16\pi m_D^3}\left|V_{ud}^*V_{cs}F_\pi f_+(0)\right|^2.}
$$
This coefficient uses exactly the stated <pion decay constant> convention and the stated massless-<pion> approximation.
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