The $W$ states are given by
(1)a)
$P_e$ is the minimum number of qubits to measure to ensure the resulting state is unentangled.
Let $P_e(\ket{\psi}) = n$ and $P_e(\ket{\phi}) = m$.
Assume $n$ qubits are measured in $\ket{\psi}$ and $m$ qubits are measured in $\ket{\phi}$. Then both $\ket{\psi}$ and $\ket{\phi}$ are unentangled and hence, $\ket{\psi} \otimes \ket{\phi}$ is also unentangled. This shows that
If $P_e(\ket{\psi} \otimes \ket{\phi}) < P_e(\ket{\psi}) + P_e(\ket{\phi})$, then either there exists a number $n' < n$ of qubits in $\ket{\psi}$ which upon measurement will ensure $\ket{\psi}$ is unentangled or there exists a number $m' < m$ of qubits in $\ket{\phi}$ which upon measurement will ensure $\ket{\phi}$ is unentangled. This is in conflict with the minimality of $n$ and $m$, hence we have equality:
(3)b)
$\ket{W_2} = \ket{\Psi^+} = \frac1{\sqrt 2}(\ket{01} + \ket{10})$ requires measurement of one qubit to ensure an unentangled state.
Write
We need measurement of $n-2$ qubits in $\ket{W_{n-1}}$ to get $\ket{W_{n-1}}$ untangled and one further measurement to ensure the sum is unentangled. Hence, $P_e(\ket{W_n}) = n-1$.
An intuitive way to get to the same result:
If a 1 is measured on a qubit in $\ket{W_n}$, then the state is no longer entangled, but if a 0 is measured, it is still entangled.
Since the maximum number of zeroes in a row that can be measured on qubits in $\ket{W_n}$ is $n-1$, we may have to do $n-1$ measurements to be certain of having an unentangled state. Hence, the persistency of $\ket{W_n}$ is $n-1$.