Pick $x_0$ as projection of $x$ onto $L$. Since $\beta^\ast$ and $x - x_0$ are parallel, ${\beta^\ast}^T(x - x_0) = \lVert \beta^\ast \rVert \lVert x - x_0 \rVert \cos\theta = \pm \lVert x - x_0 \rVert$.
$-\sum_i \alpha_i [y_i(x_i^T \beta + \beta_0) - 1] = -\sum_i \alpha_i y_i x_i^T \beta - \sum_i \alpha_i y_i \beta_0 + \sum_i \alpha_i$, which is $-\sum_i \alpha_i y_i x_i^T \beta + \sum_i \alpha_i$ from (4.51).
Adding $\frac{1}{2} \lVert \beta \rVert^2 = \frac{1}{2} (\sum_i \alpha_i y_i x_i)^T \beta$ from (4.50), we obtain $-\frac{1}{2} \sum_i \alpha_i y_i x_i^T \beta + \sum_i \alpha_i$.
Substituting $\beta$ again, $-\frac{1}{2} (\sum_i \alpha_i y_i x_i)^T(\sum_k \alpha_k y_k x_k) + \sum_i \alpha_i = -\frac{1}{2} \sum_i \sum_k \alpha_i \alpha_k y_i y_k x_i^T x_k + \sum_i \alpha_i$.