Chọn phương án C.

• $A\in(P)\Rightarrow a\cdot1+b\cdot0+c\cdot0+d=0\Rightarrow a+d=0$ (1)
• $B\in(P)\Rightarrow a\cdot0+b\cdot1+c\cdot0+d=0\Rightarrow b+d=0$ (2)

Từ (1) và (2) suy ra $\begin{cases}
b=a\\ d=-a.
\end{cases}$

Mặt khác $\begin{aligned}[t]
\mathrm{d}\big(O,(P)\big)=\dfrac{2}{3}&\Leftrightarrow\dfrac{|a\cdot0+b\cdot0+c\cdot0+d|}{\sqrt{a^2+b^2+c^2}}=\dfrac{2}{3}\\
&\Leftrightarrow\dfrac{|d|}{\sqrt{2a^2+c^2}}=\dfrac{2}{3}\\
&\Leftrightarrow\sqrt{2a^2+c^2}=\dfrac{3}{2}|d|=\dfrac{3}{2}|-a|.
\end{aligned}$

Khi đó $\begin{aligned}[t]
\mathrm{d}\big(C,(P)\big)&=\dfrac{|-3a+b+0+d|}{\sqrt{2a^2+c^2}}\\
&=\dfrac{|-3a|}{\dfrac{3}{2}|-a|}=2.
\end{aligned}$