(2pq2)4 x 5p sqrt(96)

Apply the exponent rule $(abc)^n = a^n b^n c^n$ and $(x^m)^n = x^{mn}$.

$$(2PQ^2)^4 = 2^4 P^4 (Q^2)^4 = 16 P^4 Q^{2 \times 4} = 16 P^4 Q^8$$

Find the largest perfect square factor of 96. $96 = 16 \times 6$.

$$\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}$$

The original expression is $(2PQ^2)^4 \times 5P\sqrt{96}$.
Substitute the simplified terms from Step 1 and Step 2.

$$16 P^4 Q^8 \times 5P \times 4\sqrt{6}$$

Multiply the numerical coefficients: $16 \times 5 \times 4 = 80 \times 4 = 320$.
Multiply the $P$ terms: $P^4 \times P = P^{4+1} = P^5$.
The $Q$ term is $Q^8$.
The radical term is $\sqrt{6}$.

Combine all parts.

$$320 P^5 Q^8 \sqrt{6}$$

The final answer is $\boxed{320 P^5 Q^8 \sqrt{6}}$.