= Solution
A <closed migration process> is a <continuous-time Markov chain> with state $n=(n_1,\ldots,n_m)\in\mathbb Z_+^m$, conserved population $\sum_i n_i=N$, and transition rates
$$
q(n,n-e_i+e_j)=\phi_i(n_i)p_{ij}\qquad(i\ne j).
$$
Here $\phi_i(0)=0$, $\phi_i(k)>0$ for $k>0$, and $P=(p_{ij})$ is a fixed stochastic routing <matrix>. Take $p_{ii}=0$ for convenience; self-routing can instead be omitted from the actual jumps. Assume irreducible routing. Let positive traffic weights satisfy $\theta_j=\sum_i\theta_i p_{ij}$. Define $d_i(0)=1$, $d_i(k)=\prod_{h=1}^k\phi_i(h)$. The <product-form stationary distribution of a closed migration process> is
$$
\boxed{\pi(n)=\frac1{G_N}\prod_i\frac{\theta_i^{n_i}}{d_i(n_i)},\qquad G_N=\sum_{\sum_i n_i=N}\prod_i\frac{\theta_i^{n_i}}{d_i(n_i)}.}
$$
To prove it, an incoming transition $i\to j$ to $n$ comes from $m=n+e_i-e_j$, and is possible only if $n_j>0$. The weight ratio gives
$$
\pi(m)\phi_i(n_i+1)p_{ij}=\pi(n)\frac{\theta_i}{\theta_j}\phi_j(n_j)p_{ij}.
$$
Summing over $i$ and using the traffic equations gives $\pi(n)\phi_j(n_j)$. Summing then over $j$ equals $\pi(n)\sum_j\phi_j(n_j)$, the outgoing probability flux, proving <global balance for a continuous-time Markov chain>. The state space is finite and irreducible, so this normalized <stationary distribution> is unique. <Detailed balance> is not required; a directed routing cycle, for example, is not reversible.
For the switchboard, make each of the $N$ lines one circulating individual. Use colonies $F,O,C$ for free lines, calls waiting for or receiving the operator's connection service, and already connected calls. The transitions form the cycle $F\to O\to C\to F$, with departure rates
$$
\phi_F(f)=\nu\mathbf1_{\{f>0\}},\qquad\phi_O(k)=\lambda\mathbf1_{\{k>0\}},\qquad\phi_C(i)=\mu i.
$$
The first rate is $\nu$, not $f\nu$: incoming calls form one external <Poisson process>, not one per free line. An arrival when $f=0$ changes no state and is counted as lost. Exponential connection and conversation times make this the required <continuous-time Markov chain>; only one call receives the operator's service at a time, whereas every connected call independently completes at rate $\mu$.
The cyclic traffic weights can all be chosen equal to one. Multiplying all unnormalized weights by the same $\nu^N$ gives, for $f+k+i=N$,
$$
\pi(f,k,i)\propto\left(\frac\nu\lambda\right)^k\frac{(\nu/\mu)^i}{i!}.
$$
If $n=k+i$ is the number of occupied lines, sum over the $n+1$ possibilities for the connected-call count:
$$
\sum_{i=0}^n\left(\frac\nu\lambda\right)^{n-i}\frac{(\nu/\mu)^i}{i!}=\left(\frac\nu\lambda\right)^n\sum_{i=0}^n\frac{(\lambda/\mu)^i}{i!}=H(n).
$$
Thus the normalizer is $\sum_{n=0}^NH(n)$ and the full-state probability is $H(N)/\sum_{n=0}^NH(n)$. By <Poisson arrivals see time averages>, this is also the long-run fraction of attempted calls rejected:
$$
\boxed{P_{\rm loss}=\frac{H(N)}{\sum_{n=0}^NH(n)}.}
$$
One can also see the arrival-rate argument directly: the lost-call counting process has conditional intensity $\nu\mathbf1_{\{F=0\}}$, so its equilibrium rate is $\nu\pi(F=0)$; dividing by the total attempted-call rate $\nu$ gives the same fraction. As an independent check, <stationary flow conservation in a finite-line switchboard> gives $\nu(1-P_{\rm loss})=\lambda\Pr\{O>0\}=\mu\mathbb E[C]$.
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