Answer
$v = \sqrt{\frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}}$
Work Step by Step
The kinetic energy of block A, block B, and the pulley will be equal in magnitude to the change in potential energy of block B minus the magnitude of work done by friction.
$KE = PE + work$
$\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I\omega^2 = m_Bgd - m_Ag~\mu_k~d$
$\frac{1}{2}m_Av^2+\frac{1}{2}m_Bv^2+\frac{1}{2}I(\frac{v}{R})^2 = (m_Bg - m_Ag~\mu_k)~d$
$v^2 = \frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}$
$v = \sqrt{\frac{(2)(m_Bg - m_Ag~\mu_k)~d}{m_A+m_B+\frac{I}{R^2}}}$
