= Solution
Since $M\succeq0$ but $M\notin S$, its eigenvalues satisfy
$$
\sum_{i=1}^d\mu_i=\operatorname{Tr}M>s.
$$
The continuous decreasing function
$$
q(\rho)=\sum_{i=1}^d(\mu_i-\rho)_+
$$
falls from $q(0)>s$ to zero, so there is a unique $\rho>0$ for which $q(\rho)=s$.
Put
$$
p_i=(\mu_i-\rho)_+,
\qquad
\Pi=\sum_{i=1}^dp_iv_iv_i^T.
$$
Then $\Pi\succeq0$ and $\operatorname{Tr}\Pi=\sum_i p_i=s$, so $\Pi\in S$. Moreover
$$
A:=M-\Pi
=\sum_i\min(\mu_i,\rho)v_iv_i^T,
$$
and therefore
$$
0\preceq A\preceq\rho I.
$$
For any $Z\in S$,
$$
\operatorname{Tr}(AZ)\leq\rho\operatorname{Tr}Z\leq\rho s.
$$
Whenever $p_i>0$, the corresponding eigenvalue of $A$ is exactly $\rho$, while terms with $p_i=0$ contribute nothing. Consequently
$$
\operatorname{Tr}(A\Pi)=\rho\sum_i p_i=\rho s.
$$
It follows that
$$
\langle M-\Pi,Z-\Pi\rangle_F
=\operatorname{Tr}(AZ)-\operatorname{Tr}(A\Pi)\leq0.
$$
Part (a), equivalently the formula for <projection onto a positive semidefinite trace ball>, now gives
$$
\boxed{\pi(M)=\sum_{i=1}^d
\max(0,\mu_i-\rho)v_iv_i^T},
\qquad
\sum_i\max(0,\mu_i-\rho)=s.
$$
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