In "Convex Optimization" of Boyd, a saddle-point $(v_0, x_0)$ for a Lagrangian L is the one that satisfies the below inequalities, with $v$ and $x$ as variables of the dual and primal problems respectively (assuming the primal problem does not have equality constraint),
$$L(v_0, x) <= L(v_0, x_0) <= L(v, x_0) \space | \space \forall (x,v) $$
The optimal values of the dual and primal problem can be represented as the two expressions repspectively, $$ d^* = \sup_v \inf_x L(v,x) \\ p^* = \inf_x \sup_v L(v, x) $$
Strong duality holds, if $$ p^* = d^* $$
How to prove the existence of the saddle point implies the below equations? $$ p^* = d^* = L(v_0, x_0) $$
Suppose $(v_0, x_0)$ is a saddle point. Note that \begin{align*} \forall x, L(v_0, x) \le L(v_0, x_0) &\implies \sup_x L(v_0, x) = L(v_0, x_0) \\ &\implies \inf_v \sup_x L(v, x) \le L(v_0, x_0). \end{align*} Suppose the inequality is strict. Then there exists some $v_1$ such that $$\sup_x L(v_1, x) < L(v_0, x_0).$$ But then, $$L(v_0, x_0) \le L(v_1, x_0) \le \sup_x L(v_1, x) < L(v_0, x_0).$$ This is a contradiction, hence $p^* = L(v_0, x_0)$.
We can show the other equality similarly, or simply by replacing $L$ with $-L$, and swapping the roles of $v$ and $x$.