Chọn phương án B.

Với mọi \(x,\,y\) không âm ta có $$\begin{aligned}
2x+y\cdot4^{x+y-1}\geq3\Leftrightarrow&x+y\dfrac{4^{x+y-\tfrac{3}{2}}}{2}\geq\dfrac{3}{2}\\
\Leftrightarrow&x+y\cdot4^{x+y-\tfrac{3}{2}}\geq\dfrac{3}{2}\\
\Leftrightarrow&x+y-\dfrac{3}{2}+y\cdot4^{x+y-\tfrac{3}{2}}-y\geq0\\
\Leftrightarrow&\left(x+y-\dfrac{3}{2}\right)+y\left(4^{x+y-\tfrac{3}{2}}-1\right)\geq0\;(1)
\end{aligned}$$
Nếu \(x+y-\dfrac{3}{2}< 0\) thì $$\left(x+y-\dfrac{3}{2}\right)+y\left(4^{x+y-\dfrac{3}{2}}-1\right)<0\;\text{(vô lí)}$$Vậy \(x+y\ge\dfrac{3}{2}\).

\(\begin{aligned}\text{Ta có  }P&=x^2+y^2+4x+6y\\ &=\left(x+3\right)^2+\left(y+2\right)^2-13.\end{aligned}\)

Áp dụng bất đẳng thức Bunhyakovski ta được $$\begin{eqnarray*}
&\left(1^2+1^2\right)\left(\left(x+3\right)^2+\left(y+2\right)^2\right)&\geq\left(1(x+3)+1(y+2)\right)^2\\
\Leftrightarrow&2\left(\left(x+3\right)^2+\left(y+2\right)^2\right)&\geq\left(x+y+5\right)^2\\
\Leftrightarrow&\left(x+3\right)^2+\left(y+2\right)^2&\geq\dfrac{1}{2}\left(x+y+5\right)^2\\
\Leftrightarrow&\left(x+3\right)^2+\left(y+2\right)^2-13&\geq\dfrac{1}{2}\left(x+y+5\right)^2-13\\
\Leftrightarrow&P&\geq\dfrac{1}{2}\left(\dfrac{3}{2}+5\right)^2-13\\
\Leftrightarrow&P&\geq\dfrac{65}{8}.
\end{eqnarray*}$$
Đẳng thức xảy ra khi $$\begin{cases}
x+y=\dfrac{3}{2}\\
x+3=y+2
\end{cases}\Leftrightarrow\begin{cases}
y=\dfrac{5}{4}\\
x=\dfrac{1}{4}.
\end{cases}$$