Преобразуем выражение:
$$\frac{64b^2+128b+64}{b} : \left(\frac{4}{b}+4\right) = \frac{64(b^2+2b+1)}{b} : \left(\frac{4+4b}{b}\right) = \frac{64(b+1)^2}{b} \cdot \frac{b}{4(1+b)} = \frac{64(b+1)^2}{4(b+1)} = 16(b+1)$$Подставим $$b = -\frac{15}{16}$$:
$$16\left(-\frac{15}{16}+1\right) = 16\left(-\frac{15}{16}+\frac{16}{16}\right) = 16 \cdot \frac{1}{16} = 1$$Ответ: 1