= Solution
Use the intended finite-dimensional <irreducible> <highest weight> <module> $A^\phi$, with dominant integral <highest weight> $\phi$. Its <highest weight> multiplicity is one. An arbitrary <direct sum> having that same largest <representation weight> would not in general satisfy this formula. Set $f=\pi_{A^\phi}$ and use the corrected denominator $D$ of part (a).
Restriction to each <simple root> $\mathfrak{sl}_2$ decomposes into symmetric finite <representation weight> strings, so $f$ is Weyl-invariant. Consequently $Df$ is Weyl-antisymmetric. We prove that no lower Weyl alternants occur, rather than assuming the desired character formula.
First establish the denominator identity itself. The finite sum $D$ has support in $\rho-Q_+$ and coefficient one at $\rho$. Antisymmetry kills all singular <representation weight> coefficients, since a <reflection> fixing the <representation weight> reverses their sign. Every remaining Weyl orbit has a unique regular dominant representative $\eta$. If it occurs in $D$, then $\eta\le\rho$. Since $\eta$ is integral and regular dominant, $\eta-\rho$ is dominant integral; it also belongs to $-Q_+$. A dominant vector $v=-\sum n_i\alpha_i$ satisfies $(v,v)=-\sum n_i(v,\alpha_i)\le0$, so it must be zero. Thus the only possible orbit is $W\rho$. The leading coefficient gives
$$
D=\sum_{w\in W}(\det w)e^{w\rho}.
$$
For the formal <Laplacian> $\mathscr D(e^\lambda)=\|\lambda\|^2e^\lambda$, this implies $\mathscr D D=\|\rho\|^2D$.
Now choose <root> vectors dual under an invariant form, so $[E_\alpha,E_{-\alpha}]=H_\alpha$ with $H_\alpha$ corresponding to the vector $\alpha$, and take the quadratic Casimir
$$
C=\sum_iH_i^2+\sum_{\alpha>0}(E_\alpha E_{-\alpha}+E_{-\alpha}E_\alpha)
$$
for an orthonormal Cartan <basis>. On a highest vector its scalar is $c_\phi=\|\phi+\rho\|^2-\|\rho\|^2$, hence it has this scalar throughout the cyclic <highest weight> <module>.
Here is the computation of the weighted <trace> of its radial identity. For a <positive root>, let $X$ be the formal weighted <trace> of $E_{-\alpha}E_\alpha$ and $Y$ that of $E_\alpha E_{-\alpha}$. Cyclicity of <trace> and <representation weight> shifting give $Y=e^\alpha X$, while their difference is the weighted <trace> of $H_\alpha$, namely $(\alpha,\nabla f)$. Thus
$$
X+Y=\frac{1+e^{-\alpha}}{1-e^{-\alpha}}(\alpha,\nabla f).
$$
Taking the <trace> of $C$ yields
$$
c_\phi f=\mathscr D f+\sum_{\alpha>0}\frac{1+e^{-\alpha}}{1-e^{-\alpha}}(\alpha,\nabla f).
$$
The product expression for $D$ gives
$$
\nabla\log D=\rho+\sum_{\alpha>0}\frac{e^{-\alpha}}{1-e^{-\alpha}}\alpha.
$$
Using $2\rho=\sum_{\alpha>0}\alpha$ and the <product rule> for the formal <Laplacian> therefore gives
$$
\mathscr D(Df)=c_\phi Df+f\mathscr DD
=\|\phi+\rho\|^2Df.
$$
Hence every <representation weight> in the support of $Df$ has <norm> $\|\phi+\rho\|$. Its support is also contained in $\phi+\rho-Q_+$, and it is antisymmetric. Any occurring regular <dominant weight> $\eta$ thus satisfies $\eta\le\phi+\rho$. If the inequality were strict, then
$$
\|\phi+\rho\|^2-\|\eta\|^2
=(\phi+\rho-\eta,\phi+\rho+\eta)>0,
$$
because the first factor is a nonzero nonnegative sum of simple roots and the second is strictly dominant. This contradicts the equal-<norm> condition. Only the orbit $W(\phi+\rho)$ remains. Its leading coefficient is one, proving
$$
\boxed{D*\pi_{A^\phi}=\sum_{w\in W}(\det w)\delta_{w(\phi+\rho)}.}
$$
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