b) It is not true.

a) Without loss of generality we assume that $\alpha = \angle BAD \le 90^\circ$. In this case we have
$$\begin{aligned} BD &= \sqrt{AB^2 + AD^2 - 2AB \cdot AD \cos \alpha} \le \\ &\le \sqrt{AB^2 + AD^2 + 2AB \cdot AD \cos \alpha} = [AD = BC] = \\ &= \sqrt{AB^2 + BC^2 - 2AB \cdot BC \cos(\pi - \alpha)} = AC. \end{aligned}$$
So, it is sufficient to show that $XY + YZ \ge BD$. Let $YM \parallel AD$, $DM \parallel XY$ (see the Fig.). It is evident that $XYMD$ is a parallelogram, so $YM = XD$ and $XY = MD$. Since $CZ = AX$, we have $XD = AD - AX = BC - CZ = ZB$. Since $ZB \parallel XD \parallel YM$, we see that $YBZM$ is a parallelogram.

In the parallelogram $YBZM$ we have $\angle BYM = \angle BAD = \alpha \le 90^\circ$, therefore,
$$\begin{aligned} YZ &= \sqrt{YB^2 + BZ^2 - 2YB \cdot BZ \cos(\pi - \alpha)} \ge [BZ = YM] \ge \\ &\ge \sqrt{YB^2 + YM^2 - 2YB \cdot YM \cos \alpha} = BM. \end{aligned}$$
Thus $XY + YZ \ge XY + BM = MD + BM \ge BD$, as required.

b) We show that there exist a parallelogram $ABCD$ and a point $Y$ such that the inequality
$$XY + YZ \ge \frac{AC + BD}{2} \quad (*)$$
is not valid. Let a parallelogram $ABCD$ is different from a rectangle. Then its diagonals are not equal. Without loss of generality we assume that $AC > BD$. Let $YZ \parallel AC$ (see the Fig.). Let $AX/AD = CZ/BC = \lambda$. By Thales' theorem, we have $AY/AB = CZ/BC = \lambda$. Hence $AY/AB = AX/AD = \lambda$. It follows that the triangles $AYX$ and $ABD$ are similar and $XY/BD = \lambda$, i.e. $XY = \lambda BD$.

Moreover, by construction of $Y$, the triangles $YBZ$ and $ABC$ are similar and $YZ/AC = BZ/BC = (BC - CZ)/BC = 1 - \lambda$, hence $YZ = (1 - \lambda)AC$.

Therefore, $XY + YZ = \lambda BD + (1 - \lambda)AC$ and (*) has the form $\lambda BD + (1 - \lambda)AC \ge \frac{1}{2}BD + \frac{1}{2}AC$ or $(\lambda - \frac{1}{2})BD \ge (\lambda - \frac{1}{2})AC$. But this inequality is not valid if $\lambda > \frac{1}{2}$.