Proof
Notice, by the definition of conditional probability 4.5.3 and the multiplication rule 4.5.5
\begin{equation*}
P(S_j | A) = \frac{P(S_j \cap A)}{P(A)} = \frac{P(S_j)P( A | S_j)}{P(A)}.
\end{equation*}
But using the disjointness of the partition
\begin{align*}
P(A) & = P( (A \cap S_1) \cup (A \cup S_2) \cup ... \cup (A \cup S_m) )\\
& = P(A \cap S_1) + P(A \cup S_2) + ... + P(A \cup S_m)\\
& = P(S_1 \cap A) + P(S_2 \cup A) + ... + P(S_m \cup A)\\
& = P(S_1) P(A | S_1) + P(S_2)P(A | S_2) + ... + P(S_m)P(A | S_m)\\
& = \sum_{k=1}^m P(S_k)P(A | S_k)
\end{align*}
Put these two expansions together to obtain the desired result.