If I understand correctly, you have a linear programming problem $P$ and a basic solution $x^*$ with corresponding basic solution $y^*$ of the dual problem $D$ such that $y^*$ is feasible for $D$ but $x^*$ is not feasible for $P$ due to a single violated inequality constraint $C_i$.
Of course it's possible that there are no optimal solutions at all. It is also possible that there are optimal solutions in which your constraint $C_i$ is not an equality.
Consider the rather trivial linear programming problem $P$:
maximize $0$
subject to $$ \eqalign{x_1 &\le 1\cr x_1 &\le 2\cr x_1 &\ge 0\cr} $$ Let $s_1$ and $s_2$ be the slack variables corresponding to the two constraints. The basic solution for basis $x_1, s_1$ is $x_1 = 2$, $s_1 = -1$, $s_2 = 0$ which violates the first constraint. All basic solutions of the dual are $y_1 = 0$, $y_2 = 0$ which is feasible. There are optimal solutions $0 \le x_1 \le 1$.
EDIT: We can indeed say that if $P$ has an optimal solution, it has an optimal solution in which $C_i$ is an equality.
Proof: Suppose $x^{**}$ is an optimal solution of $P$ in which constraint $C_i$ is not an equality. Then an appropriate convex combination $x^{***}$ of $x^*$ (in which the slack variable corresponding to $C_i$ has a negative value) and $x^{**}$ (where it has a positive value) will have $C_i$ satisfied as an equality. Since both $x^{*}$ and $x^{**}$ satisfy all the other constraints, $x^{***}$ is feasible for $P$.
Now $x^*$ is an optimal solution for $P_i$: note that the dual variable value $y^*_i$ for the omitted constraint $C_i$ in $y^*$ must be $0$ by complementary slackness, so $y^*$ (with the $i$ component omitted) is still feasible for the dual of $P_i$. On the other hand, $x^{**}$ is feasible for $P_i$. Thus if $f$ is the objective function, $f(x^*) \ge f(x^{**})$, and $f(x^{***}) \ge f(x^{**})$. Thus $x^{***}$ is an optimal solution for $P$.