= Solution
Let $\Gamma=T_B$ denote the <partial transpose>. The <partial transpose of a maximally entangled projector> is
$$
(\phi^+)^\Gamma=\frac1d\sum_{i,j}|i\rangle\langle j|\otimes|j\rangle\langle i|=\frac Sd,
$$
where $S$ is the <swap operator>, satisfying $S^\dagger S=I$ and $\|S\|_{\rm op}=1$. If $\sigma$ has <positive partial transpose>, then $\sigma^\Gamma$ is a <positive operator> of <trace> one, hence $\|\sigma^\Gamma\|_1=1$.
For a <pure state> $\phi^+=|\Phi\rangle\langle\Phi|$, the unsquared <quantum fidelity> obeys $F(\phi^+,\sigma)^2=\langle\Phi|\sigma|\Phi\rangle$. Using the <trace-adjoint identity for partial transpose> and part (i),
$$
F(\phi^+,\sigma)^2
=\operatorname{Tr}(\phi^+\sigma)
=\frac1d\operatorname{Tr}(S\sigma^\Gamma)
\leq\frac1d\|\sigma^\Gamma\|_1=\frac1d.
$$
The overlap is nonnegative, so taking its square root proves \b[the <PPT maximally entangled overlap bound>]:
$$
\boxed{F(\phi^+,\sigma)\leq\frac1{\sqrt d}.}
$$
The <product state> $|00\rangle\langle00|$ attains the bound, so it is sharp.
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