7. Given the real-coefficient equation $a x^{3}-x^{2}+b x-1=0$ has three positive real roots. Then
$$
P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)}
$$

the minimum value of $P$ is
7. 108.

Let the three positive real roots of $a x^{3}-x^{2}+b x-1=0$ be $v_{1}, v_{2}, v_{3}$.
By Vieta's formulas, we have
$$
\begin{array}{l}
v_{1}+v_{2}+v_{3}=\frac{1}{a}, \\
v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}=\frac{b}{a}, \\
v_{1} v_{2} v_{3}=\frac{1}{a} .
\end{array}
$$

From (1) and (2), we get $a>0, b>0$.
From (1) and (3), we get
$$
\frac{1}{a} \geqslant 3 \sqrt{3}.
$$

And $3\left(v_{1} v_{2}+v_{2} v_{3}+v_{3} v_{1}\right) \leqslant\left(v_{1}+v_{2}+v_{3}\right)^{2}$
$$
\Rightarrow 3 \cdot \frac{b}{a} \leqslant \frac{1}{a^{2}} \Rightarrow 3 a b \leqslant 1.
$$

Thus, $P=\frac{5 a^{2}-6 a b+3}{a^{3}(b-a)} \geqslant \frac{5 a^{2}+1}{a^{3}(b-a)}$.
Also, $\left(v_{2}+v_{3}-v_{1}\right)\left(v_{3}+v_{1}-v_{2}\right)\left(v_{1}+v_{2}-v_{3}\right)$
$\leqslant v_{1} v_{2} v_{3}$
$\Rightarrow\left(\frac{1}{a}-2 v_{1}\right)\left(\frac{1}{a}-2 v_{2}\right)\left(\frac{1}{a}-2 v_{3}\right) \leqslant \frac{1}{a}$
$\Rightarrow 9 a^{2}-4 a b+1 \geqslant 0$
$\Rightarrow 5 a^{2}+1 \geqslant 4 a(b-a)$.
Then $P \geqslant \frac{4 a(b-a)}{a^{3}(b-a)}=\frac{4}{a^{2}} \geqslant 108$.
Therefore, when $a=\frac{\sqrt{3}}{9}, b=\sqrt{3}$, $P$ achieves its minimum value of 108.