As the top boundary curve $P(u, 1)$ is defined by 3 points, we select the quadratic curve bezier.

$P(u, 1)=(1-u)^{2} P_{4}+2(1-u) P_{5}+u^{2} P_{6}$

$P_{x}(u, 1)=(1-u)^{2}(-1)+2 u(1-u)(0)+u^{2}(1)$

$\quad \quad \quad \quad=2 u-1$

$\begin{aligned} P_{y}(u, 1) &=(1-u)^{2}(0)+2 u(1-u)(-1)+u^{2}(0) \\ &=-2 u(1-u) \\ P_{z}(u, 1) &=(1-u)^{2}(1)+2 u(1-u)(1)+u^{2}(1) \\ &=1 \end{aligned}$

$\therefore P(u, v)=\left[\begin{array}{llll}{2 u-1} & {-2 u(1-u)} & {1}\end{array}\right]$

Similarly,

$P(u, 0)=(1-u)^{2} P_{1}+2(1-u) P_{2}+u^{2} P_{3}$

$P_{x}(u, 0)=(1-u)^{2}(-1)+2 u(1-u)(0)+u^{2}(1)$

$\quad \quad \quad \quad=2 u-1$

$P_{y}(u, 0)=(1-u)^{2}(0)+2 u(1-u)(-1)+u^{2}(0)$

$\quad \quad \quad \quad =-2 u(1-u)$

$\begin{aligned} P_{z}(u, 0) &=(1-u)^{2}(0)+2 u(1-u)(0)+u^{2}(0) \\ &=0 \end{aligned}$

The equation of surface,

$P(u, v)=(1-v) P(u, 0)+v P(u, 1)$

$P_{x}(u, v)=(1-v)(2 u-1)+v(2 u-1)$

$\quad \quad \quad \quad=2 u-1$

$\begin{aligned} P_{y}(u, v) &=(1-v)(-2 u(1-u))+(-2 u(1-u)) \\ &=-2 u(1-u) \\ P_{z}(u, v) &=(1-v)(0)+v(1) \\ &=v \end{aligned}$,

$\therefore$ Equation of lofted surface is,

$P(u, v)=\left[\begin{array}{ccc}{2 u-1} & {-2 u(1-u)} & {v]}\end{array}\right.$