7. Let positive real numbers $x, y$ satisfy $x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y}=\frac{27}{4}$. Then the minimum value of $P=\frac{15}{x}-\frac{3}{4 y}$ is
7. 6 .

By the AM-GM inequality for three terms, we have
$$
\begin{array}{l}
x^{2}+\frac{1}{x}=\left(x^{2}+\frac{8}{x}+\frac{8}{x}\right)-\frac{15}{x} \geqslant 12-\frac{15}{x}, \\
y^{2}+\frac{1}{y}=\left(y^{2}+\frac{1}{8 y}+\frac{1}{8 y}\right)+\frac{3}{4 y} \geqslant \frac{3}{4}+\frac{3}{4 y} .
\end{array}
$$

Adding the two inequalities, we get
$$
\begin{array}{l}
\frac{27}{4}=x^{2}+y^{2}+\frac{1}{x}+\frac{1}{y} \geqslant \frac{51}{4}+\left(\frac{3}{4 y}-\frac{15}{x}\right) \\
\Rightarrow P=\frac{15}{x}-\frac{3}{4 y} \geqslant 6 .
\end{array}
$$

Equality holds if and only if $x=2, y=\frac{1}{2}$. Therefore, the minimum value of $P$ is 6.