= Solution
The <Lagrangian sufficiency theorem> for an equality constraint states that if a feasible $x_*$ and a multiplier $\lambda_*$ exist such that $x_*$ globally maximizes $L(x,\lambda_*)=f(x)-\lambda_*\cdot(g(x)-b)$ over the underlying domain, then $x_*$ globally maximizes $f$ over $g(x)=b$. Indeed $L=f$ at every feasible point, so global maximality of $L$ directly compares the feasible objective values. The underlying domain can already impose $x_i\geq0$.
For $0<p<1$, choose $\lambda=p d^{1-p}$. Each function $t^p-\lambda t$ on $t\geq0$ is a <strictly concave function> and attains its unique maximum at $t=1/d$, since $pt^{p-1}=\lambda$ there. The sum <Lagrangian> is therefore globally maximized when every $x_i=1/d$, and this point is feasible. For $p=1$ the objective is identically one on the feasible set. For $p>1$, each feasible $x_i$ lies in $[0,1]$, so $x_i^p\leq x_i$, with equality only at $0$ or $1$; equality in the total is attained exactly at a <simplex> vertex. Consequently
$$
\boxed{\max\sum_{i=1}^d x_i^p=\begin{cases}d^{1-p},&0<p<1,\\1,&p\geq1.\end{cases}}
$$
For $0<p<1$ the unique maximizer is $(1/d,\ldots,1/d)$; for $p=1$ every feasible point is a maximizer; for $p>1$ the maximizers are exactly the $d$ vertices with one coordinate one and all others zero. When $d=1$ these descriptions coincide.
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