$$\frac{(3-2i^{3})(2+3P)}{(9+2i)(2-i^{3})}$$
$\left(\frac{222}{425}-\frac{21}{425}i\right)P+\left(\frac{148}{425}-\frac{14}{425}i\right)$
$$\frac{\left(3-2\left(-i\right)\right)\left(2+3P\right)}{\left(9+2i\right)\left(2-i^{3}\right)}$$
$$\frac{\left(3-\left(-2i\right)\right)\left(2+3P\right)}{\left(9+2i\right)\left(2-i^{3}\right)}$$
$$\frac{\left(3+2i\right)\left(2+3P\right)}{\left(9+2i\right)\left(2-i^{3}\right)}$$
$$\frac{\left(3+2i\right)\left(2+3P\right)}{\left(9+2i\right)\left(2-\left(-i\right)\right)}$$
$$\frac{\left(3+2i\right)\left(2+3P\right)}{\left(9+2i\right)\left(2+i\right)}$$
$$\frac{\left(3+2i\right)\left(2+3P\right)}{16+13i}$$
$$\left(\frac{74}{425}-\frac{7}{425}i\right)\left(2+3P\right)$$
$$\frac{148}{425}-\frac{14}{425}i+\left(\frac{222}{425}-\frac{21}{425}i\right)P$$
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