(Ⅰ) We start from the equation $S_n = npa_n - np + n$,

For $n = 1$, we have $S_1 = a_1 = pa_1 - p + 1$,
which simplifies to $(1 - p)(1 - a_1) = 0$.
Thus, we get $p = 1$ or $a_1 = 1$.

For $n = 2$, we have $S_2 = a_1 + a_2 = 2pa_2 - 2p + 2$,
If $p = 1$, then $a_1 + a_2 = 2a_2 - 2 + 2 = 2a_2$, and this implies $a_1 = a_2$, which contradicts our initial condition that $a_1 \neq a_2$. Therefore, $p \neq 1$,
and it follows that $a_1 = 1$. Since $a_1 \neq a_2$, we also have $a_2 \neq 1$.
Substituting $a_1 = 1$ into the equation for $n = 2$, we solve for $p$ and obtain $\boxed{p = \frac{1}{2}}$.

(Ⅱ) Substituting $p = \frac{1}{2}$ into $S_n = npa_n - np + n$, we get $2S_n = n(a_n + 1)$.
When $n \geq 2$, let's consider the sum of first $(n - 1)$ terms: $2S_{n-1} = (n - 1)(a_{n-1} + 1)$.

Subtract the second equation from the first one to get:
$$2S_n - 2S_{n-1} = n(a_n + 1) - (n - 1)(a_{n-1} + 1).$$

The left-hand side simplifies to $2a_n$ and the right-hand side simplifies to $(n - 2)a_n - (n - 1)a_{n-1} + 1$, which gives us
$$(n - 2)a_n - (n - 1)a_{n-1} + 1 = 0.$$

Similarly, the next term of the sequence satisfies:
$$(n - 1)a_{n+1} - na_n + 1 = 0.$$

Subtracting these two latest equations gives us:
$$a_{n+1} - 2a_n + a_{n-1} = 0,$$

which implies that $\boxed{a_{n+1} - a_n = a_n - a_{n-1}}$ for $n \geq 2$.

This condition shows that the differences between consecutive terms of the sequence are constant, which is the characteristic property of an arithmetic sequence. Therefore, the sequence $\{a_n\}$ is indeed an arithmetic sequence.