Given

$\rm \frac{b}{x - 2} + \frac{x + 3}{2 - x} = \frac{1 - x}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} - \frac{x + 3}{2 - x}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} - \frac{x + 3}{ - (x - 2)}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x}{x - 2} + \frac{x + 3}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{ (1 - x) + (x + 3)}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{1 - x + x + 3}{x - 2}$

$\rm or, \frac{b}{x - 2} = \frac{4}{x - 2}$

The expressions in the denominators of both sides of the equation are equal. So, the numerators of these expressions must be equal to hold the equality.

$\rm \therefore b = 4$

Hence, the required value of b is 4.