= Solution
Write $h(t)=t/(1+t)$ for $t\ge0$. Every summand in $d$ is nonnegative and at most $2^{-m}$, so the series converges. Symmetry is immediate. If $d(x,y)=0$, then $\phi_m(x-y)=0$ for every $m$. For arbitrary $\phi\in(\ell^p)^*$, choose a sequence of $\phi_m$ converging to $\phi$ in operator norm. Since $x-y$ is fixed, $\phi(x-y)=0$. The <continuous dual space> separates points, so $x=y$.
The function $h$ is increasing and subadditive:
$$
h(a+b)\le h(a)+h(b)\qquad(a,b\ge0).
$$
Combining this with the <triangle inequality> for each $|\phi_m(x-y)|$ and summing proves the triangle inequality for $d$. Hence $d$ is a <metric> on $B$.
Suppose first that $x^{(n)}$ converges weakly to $x$, meaning that
$$
\phi(x^{(n)})\to\phi(x)
\qquad\text{for every }\phi\in(\ell^p)^*.
$$
For each fixed $m$, the $m$th summand of $d(x^{(n)},x)$ tends to zero. A finite-head and geometric-tail estimate, using that every summand is bounded by $2^{-m}$, then gives $d(x^{(n)},x)\to0$.
Conversely, if $d(x^{(n)},x)\to0$, each nonnegative summand tends to zero, and therefore
$$
\phi_m(x^{(n)}-x)\to0
$$
for every $m$. Given $\phi\in(\ell^p)^*$ and $\varepsilon>0$, choose $m$ with $\|\phi-\phi_m\|<\varepsilon$. Since $x^{(n)},x\in B$,
$$
|\phi(x^{(n)}-x)|
\le|\phi_m(x^{(n)}-x)|+2\varepsilon.
$$
Taking the limit superior and then letting $\varepsilon\downarrow0$ proves weak convergence. Thus
$$
\boxed{\phi(x^{(n)})\to\phi(x)\ \forall\phi\in(\ell^p)^*
\iff d(x^{(n)},x)\to0}.
$$
The same approximation argument for neighbourhoods shows that $d$ induces the weak topology on $B$, as in the <metrization of the weak topology on a bounded set>.
If $q$ is conjugate to $p$, the given <duality of sequence spaces> applied twice gives
$$
(\ell^p)^{**}\cong(\ell^q)^*\cong\ell^p,
$$
so $\ell^p$ is a <reflexive Banach space>. Its closed unit ball is weakly compact by the <weak compactness characterization of reflexivity>. Since $d$ induces that topology on $B$,
$$
\boxed{(B,d)\text{ is a compact metric space}}.
$$
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