Given that,
$\left ( \sqrt{2} \right )^{19}3^{4}4^{2}9^{m}\cdot 8^{n} = 3^{n}\cdot 16^{m} \left ( 4\sqrt{64} \right )$
$\implies 2^{19/2}\cdot 3^{4}\cdot \left ( 2^{2} \right )^{2} \left ( 3^{2} \right )^{m} \left ( 2^{3} \right )^{n} = 3^{n} \left ( 2^{4} \right )^{m} \left ( 2^{6} \right )^{1/4}$
$\implies 2^{19/2}\cdot 3^{4}\cdot 2^{4}\cdot 3^{2m} \cdot 2^{3n} = 3^{n}\cdot 2^{4m}\cdot 2^{6/4}$
$\implies 2^{\left ( 19/2 + 4 + 3n \right )}\cdot 3^{\left ( 4 + 2m \right )} = 3^{n} \:2^{\left ( 4m + \frac{6}{4} \right )}$
Comparing both sides, we get
- $4 + 2m = n \qquad \qquad \to (i)$
- $\frac{19}{2} + 4 + 3n = 4m + \frac{3}{2} \qquad \to (ii)$
$(ii) \implies 3n+ \frac{19}{2}-\frac{3}{2}+4-4m=0$
$\implies 3n+ \frac{16}{2}+4-4m=0$
$\implies 3n+ 12-4m=0$
Put the value of $n$ from equation $(i),$ we get
$3\left ( 4+2m \right )+12-4m=0$
$\implies 12+6m+12-4m=0$
$\implies 2m + 24 =0$
$\implies 2m = -24$
$\implies m = -\frac{24}{2}$
$\boxed{m = -12}$
Correct Answer$: \text{B}$