Chọn phương án D.

\begin{eqnarray*}
&\log_2x+\log_2(2y)&\geq\log_2\left(x^2+2y\right)\\
\Leftrightarrow&\log_2(2xy)&\geq\log_2\left(x^2+2y\right)\\
\Leftrightarrow&2xy&\geq x^2+2y
\end{eqnarray*}

• Xét $0<x\leq1\Rightarrow2y\geq2xy\geq x^2+2y$
  $\Rightarrow x^2\leq0$ (vô lý)
• Xét $x>1\Rightarrow2xy\geq x^2+2y$
  $\Leftrightarrow2y(x-1)\geq x^2\geq2y\geq\dfrac{x^2}{x-1}$

Suy ra $P=x+2y\ge x+\dfrac{x^2}{x-1}$.

Xét hàm số $f(x)=x+\dfrac{x^2}{x-1}$ trên $(1;+\infty)$.
Ta có $f'(x)=\dfrac{2x^2-4x+1}{(x-1)^2}$.
Cho $f'(x)=0\Leftrightarrow\left[\begin{array}{ll}x=\dfrac{2+\sqrt{2}}{2} &\in(1;+\infty)\\ x=\dfrac{2-\sqrt{2}}{2} &\notin(1;+\infty)\end{array}\right.$
Do đó $\min\limits_{(1;+\infty)}f(x)=f\left(\dfrac{2+\sqrt{2}}{2}\right)=2\sqrt{2}+3$.

Suy ra $\begin{cases}
a=b=2\\ c=3
\end{cases}$. Vậy $a+b+c=7$.