Pull it, Mr Pulley!

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Thanks to Deeparaj Bhat for an excellent explanation.

Relevant wiki: Conservation of Energy

The change in height of block B is $\sqrt{15^2+8^2} - \sqrt{(15-9)^2+8^2} = 17-10 = 7\\\Rightarrow \Delta h = 7$

Conservation of Energy gives, $T_a + T_b = \Delta U_b$

Taking constraint relations, $s_a^2 = s_b^2+h^2$

Differentiating, $2s_ads_a = 2s_bds_b\\\dfrac{ds_a}{dt}=\dfrac{s_b}{s_a}\dfrac{ds_b}{dt} = \sin\theta\dfrac{ds_b}{dt}\\v_a = v_b\sin\theta\cdots(1)$

Hence conserving energy, $\dfrac12m_av_a^2 + \dfrac12m_bv_b^2 = m_bg\Delta h$

Using $(1)$ , $\Rightarrow v_a^2(m_a + m_b\sin^2\theta) = 2m_bg\Delta h\\\boxed{v_a = \sqrt{\dfrac{2m_bg\Delta h}{m_a + m_b\sin^2\theta}}}$

Substituting values, $v_a = 8.58856$