Why is carrier wave affected by percentage modulation?
A conventional AM signal can be written as

$$x_{AM}(t)=A_c\big(1+\alpha\; m(t)\big)\cos(\omega_ct)\tag{1}$$

where $\alpha$ is the modulation index and $m(t)$ is the message signal. In Eq. $(1)$ it is assumed that the message signal $m(t)$ is normalized, i.e., $\max_t|m(t)|= 1$. The average power of $x_{AM}(t)$ is

$$\overline{x^2_{AM}(t)}=\frac{A_c^2}{2}\left(1+\alpha^2\overline{m^2(t)}\right)\tag{2}$$

If you assume now that the power $(2)$ is fixed to some value $P$, then the total available power must divided between the carrier term and the message-bearing term in $(2)$. Equating $(2)$ with the available power $P$, and expressing the carrier term $A_c^2/2$ in terms of the modulation index and the average power of the message signal gives

$$\frac{A_c^2}{2}=\frac{P}{1+\alpha^2\overline{m^2(t)}}\tag{3}$$

In the figure in your question it looks like they assumed $\overline{m^2(t)}=\frac12$, which for a maximum modulation index $\alpha=1$ results in

$$\frac{A_c^2}{2}=\frac23 P\tag{4}$$

This value corresponds to the right-most point of the curve labeled "carrier".