We have (from the book),

$d_{1}=\frac{\ln(G_{0}/K)}{\sigma\sqrt{T}}+\frac{\sigma\sqrt{T}}{2}\,,\qquad d_{2}=d_{1}-\sigma\sqrt{T}\,.$

Hence,

$\frac{\partial P_{0}^{\rm B}(K,\sigma)}{\partial\sigma}$ $=\frac{\partial}{\partial\sigma}B_{0}\big(K\Phi(-d_{2})-G_{0}\Phi(-d_{1})\big)$
$=-B_{0}K\phi(-d_{2})\frac{\partial(d_{1}-\sigma\sqrt{T})}{\partial\sigma}+B_{0}G_{0}\phi(-d_{1})\frac{\partial d_{1}}{\partial\sigma}\,.$

Using the hint and $\phi(-d)=\phi(d)$ , by symmetry,

$\frac{\partial P_{0}^{\rm B}(K,\sigma)}{\partial\sigma}=B_{0}K\phi(d_{2})\sqrt{T}>0\,,$

as $K$ , $B_{0}$ and $\phi(d_{2})$ are strictly positive. Alternatively, we can also write

$\frac{\partial P_{0}^{\rm B}(K,\sigma)}{\partial\sigma}=B_{0}G_{0}\phi(d_{1})\sqrt{T}>0\,,$

using the equation from the hint.