= Solution
Let $D(0)=1$. The <present-extrapolated spherical-collapse barrier> is $\delta_c(z)=\delta_{\rm sc}/D(z)$, where $\delta_{\rm sc}\simeq1.686$ is the linear contrast extrapolated to spherical collapse in a matter-dominated universe. It is not the nonlinear interior overdensity $18\pi^2$. The <cosmological mass variance> $\sigma^2(M,0)$ is the variance of the present-extrapolated linear <density contrast> smoothed over a mass $M$. It is not the variance of already virialized halo densities.
For a fixed-shape smoothing filter and $P(k)=P_0k^n$, rescale the variance integral with $x=kR$:
$$
\sigma^2(R)=\frac{P_0}{2\pi^2}R^{-(n+3)}\int_0^\infty x^{n+2}|W(x)|^2dx.
$$
Thus $\sigma(M,0)=S M^{-q}$, with $q=(n+3)/6$ and $S$ a normalization carrying the appropriate mass units. For a real-space <top-hat filter>, convergence requires $-3<n<1$; a spectrum outside this interval needs physical cutoffs and does not have this unrestricted scale-free result.
If $n(M,z)dM$ counts halos per comoving volume, the <Press-Schechter halo mass function> follows by differentiating the cumulative mass fraction and converting mass fraction to number:
$$
n(M,z)=-\frac{\bar\rho_{m,0}}M\frac{\partial f(>M,z)}{\partial M}
=\sqrt{\frac2\pi}\frac{\bar\rho_{m,0}}{M^2}q\nu e^{-\nu^2/2},\qquad
\nu=\frac{\delta_c(z)}{\sigma(M,0)}.
$$
The minus sign is needed because the cumulative fraction decreases with $M$. To match the printed exponential coefficient exactly, define $M_*(z)$ by
$$
\frac{\nu^2}{2}=\left(\frac M{M_*}\right)^{2q},\qquad
\boxed{\sigma(M_*,0)=\frac{\delta_c(z)}{\sqrt2}.}
$$
This <peak-height calibration from an exponential mass-function cutoff> differs from the also-common convention $\nu(M_*)=1$. Here $\nu=\sqrt2(M/M_*)^q$, and consequently
$$
\boxed{\alpha=q-2=\frac{n-9}{6},\quad\beta=2q=\frac{n+3}{3},\quad
A(z)=\frac{2q}{\sqrt\pi}\bar\rho_{m,0}M_*^{-q}
=\frac{n+3}{3\sqrt\pi}\bar\rho_{m,0}M_*^{-(n+3)/6}.}
$$
Equivalently $A=\sqrt{2/\pi}\,q\bar\rho_{m,0}\delta_c/S$. The <scale-free halo cutoff normalization> therefore evolves with $z$; the displayed $A$ is not a redshift-independent universal constant. For physical number density, replace $\bar\rho_{m,0}$ by $\bar\rho_m(z)$ instead. Since $M_*=(\sqrt2S/\delta_c)^{1/q}$, it grows as $D^{6/(n+3)}$; in matter domination it is proportional to $(1+z)^{-6/(n+3)}$. At the collapse epoch itself, $\sigma(M_*,z)=\delta_{\rm sc}/\sqrt2$.
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