The solution for minimizing the function $\mathrm{z}=x+\mathrm{y}$ under an L.P.P. with constraints $x+\mathrm{y} \geqslant 2, x+2 \mathrm{y} \leqslant 8, \mathrm{y} \leqslant 3, x, \mathrm{y} \geqslant 0$ is

The objective function $z = x + y$ is minimized subject to the constraints $x + y \ge 2$, $x + 2y \le 8$, $y \le 3$, $x \ge 0$, and $y \ge 0$.

The feasible region is bounded and a polygon with corner points at $(0,2)$, $(2,0)$, $(8,0)$, $(2,3)$, and $(0,3)$.

Evaluating $z$ at these points: $z(0,2) = 2$, $z(2,0) = 2$, $z(8,0) = 8$, $z(2,3) = 5$, and $z(0,3) = 3$.

The minimum value is $2$, occurring at $(0,2)$ and $(2,0)$. By the Fundamental Theorem of Linear Programming, every point on the line segment between $(0,2)$ and $(2,0)$ minimizes $z$.

The feasible region is bounded, and the minimization solution is consistent with option $\boxed{C}$.