Prove the MinAveCodeLen of a product information source is less than the sum of that of the multiplicand and multiplier source?
It's so wired that again, I'm answering my own question... I have to admit I am not confident at all when confronted with constructive proof problems.

An encoding scheme for ${P}$ could be described graphically as follows: Suppose $T_A$, $T_B$ are 2 Huffman trees for ${P}_A$ and ${P}_B$ respectively. For each leaf $s_{Ai}$ in $T_A$ replace it with $T_B$, i.e.

Letting $|S_A| = q,|S_B| = q$, the average code word length of this coding scheme is
$$C = \frac{\sum l_{Ai}+l_{Bj}}{pq} = \frac{q\sum l_{Ai}+p\sum l_{Bj}}{pq} = \mathrm{MinAveCodeLen}({P}_A)+\mathrm{MinAveCodeLen}\mathbb{P}_B)$$
where $l_{Ai}$, $l_{Bj}$ are the codeword length of $s_{Ai}$ and $s_{Bj}$ in the Huffman code corresponding to $T_A$, $T_B$. Therefore,
$$\mathrm{MinAveCodeLen}({P})\leq C\leq\mathrm{MinAveCodeLen}({P}_A)+\mathrm{MinAveCodeLen}({P}_B)$$