For each start point the step word is a bijection onto $S^n$

For $v\in {\mathcal{L}}_S(P;n)$ and each $i$ with $1\le i\le n$ the difference $v(i)-v(i-1)$ lies in $S$, so $\mathrm{\Phi }(v)$ is a function $\{0,\dots ,n-1\}\to S$, that is an element of $S^n$; for $n=0$ the domain is empty and $\mathrm{\Phi }(v)$ is the empty word.

Given $w\in S^n$, the traced path $v_w$ lies in ${\mathcal{L}}_S(P;n)$ and its step word has $j$-th letter $v_w(j+1)-v_w(j)=w_j$, so $\mathrm{\Phi }(v_w)=w$; conversely, given $v\in {\mathcal{L}}_S(P;n)$ with $w=\mathrm{\Phi }(v)$, both $v$ and $v_w$ take the value $P$ at the index $0$ and both satisfy $u(i)=u(i-1)+w_{i-1}$ for $1\le i\le n$, so the set of indices at which they agree contains $0$ and contains $i$ whenever it contains $i-1$, whence they agree throughout and $v_w=v$. Thus $w\mapsto v_w$ is a two-sided inverse of $\mathrm{\Phi }$ and $\mathrm{\Phi }$ is a bijection.

Since $S$ is finite and $\{0,\dots ,n-1\}$ is finite with $n$ elements, $S^n$ is finite with $\mid S^n\mid =\mid S{\mid }^n$, and transporting along the bijection of step 2.1 gives that ${\mathcal{L}}_S(P;n)$ is finite with the same cardinality. At $n=0$ both sides are $1$, one empty path against the one empty word, and this holds also for $S=\mathrm{\varnothing }$; for $S=\mathrm{\varnothing }$ and $n\ge 1$ both sides are $0$.