How would you go about solving this pH problem?
The $K_\mathrm{b}$ value of benzylamine, the conjugated base of benzylmmoniumchloride, is given. The acid constant of benzylammoniumchloride can be inferred via the identity
$$ \mathrm{p}K_a + \mathrm{p}K_b = 14. $$
Thus, we have
$$ \mathrm{p}K_\text{b,benzylamin} =  4.65$$
$$ \mathrm{p}K_\text{a,benzylammoniumchloride} = 9.35 $$
$$ \Rightarrow K_\text{a,benzylammoniumchloride} = 10^{-9.35}$$
As furthermore
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus] \cdot [\mathrm{PhCH_2NH_2}]}{[\mathrm{PhCH_2NH_3^\oplus]}} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
and
we see that
$$ K_\text{a,benzylammoniumchloride} = \frac{[\mathrm{H}^\oplus]^2}{[\mathrm{PhCH_2NH_3^\oplus]}}$$
This leaves us with an equation with only one unknown variable and is easy to solve as $[\mathrm{PhCH_2NH_3^\oplus}] = 0.24~\mathrm{\frac{mol}{L}}$ is given:
$$
[\mathrm{H^\oplus}] = \sqrt{[\mathrm{PhCH_2NH_3^\oplus]}\cdot K_\text{a,benzylammoniumchloride}} \approx 1.0354 \cdot 10^{-5} ~\mathrm{mol/L}
$$
And as the pH is the negative of the logarithm with base ten we have:
$p\mathrm{H} \approx 4.985$