Chọn phương án B.

Ta có \(\displaystyle\int f(x)\mathrm{\,d}x=\dfrac{1}{3}ax^3+\dfrac{1}{2}bx^2+cx+d\).

• \(\displaystyle\int\limits_0^1f(x)\mathrm{\,d}x=-\dfrac{7}{2}\)
  \(\Leftrightarrow\left(\dfrac{1}{3}ax^3+\dfrac{1}{2}bx^2+cx+d\right)\bigg|_0^1=-\dfrac{7}{2}\)
  \(\Leftrightarrow\dfrac{1}{3}a+\dfrac{1}{2}b+c=-\dfrac{7}{2}\,\,(1)\)
• \(\displaystyle\int\limits_0^2 f(x)\mathrm{\,d}x=-2\)
  \(\Leftrightarrow\left(\dfrac{1}{3}ax^3+\dfrac{1}{2}bx^2+cx+d\right)\bigg|_0^2=-2\)
  \(\Leftrightarrow\dfrac{8}{3}a+2b+2c=-2\,\,(2)\)
• \(\displaystyle\int\limits_0^3 f(x)\mathrm{\,d}x=\dfrac{13}{2}\)
  \(\Leftrightarrow\left(\dfrac{1}{3}ax^3+\dfrac{1}{2}bx^2+cx+d\right)\bigg|_0^3=\dfrac{13}{2}\)
  \(\Leftrightarrow9a+\dfrac{9}{2}b+3c=\dfrac{13}{2}\,\,(3)\)

Từ \((1),\,(2),\,(3)\) ta có hệ $$\begin{cases}\dfrac{1}{3}a+\dfrac{1}{2}b+c=-\dfrac{7}{2}\\\dfrac{8}{3}a+2b+2c=-2\\ 9a+\dfrac{9}{2}b+3c=\dfrac{13}{2}\end{cases}\Rightarrow\begin{cases}a=1\\ b=3\\ c=-\dfrac{16}{3}.\end{cases}$$
Vậy \(P=a+b+c=4-\dfrac{16}{3}=-\dfrac{4}{3}\).