K_sp of PbBr₂ is 8 x 10^-5. If the salt is 80% dissociated in solution. The solubility of salt in gL^-1 is P+3.48 then the value of P is.

Let the solubility of PbBr₂ be S mol L^-1
${\mathrm{pbBr}}_{2\left(\mathrm{s}\right)}\rightleftharpoons {\mathrm{pbBr}}_{2\left(\mathrm{aq}\right)}\rightleftharpoons {\mathrm{pb}}_{\mathrm{aq}}^{+2}+2{\mathrm{Br}}_{\mathrm{aq}}^{-}$
Ionisation of pbBr₂ = 80%  $\frac{\mathrm{s}\times 80}{100};\frac{2\mathrm{s}\times 80}{100}$
$\begin{array}{l}\Rightarrow \mathrm{Ksp}=\left[{\mathrm{pb}}^{+2}\right]{\left[{\mathrm{Br}}^{-}\right]}^2\\ 8\times {10}^{-5}=\left(\frac{\mathrm{S}\times 80}{100}\right){\left(\frac{2\mathrm{S}\times 80}{100}\right)}^2\end{array}$
From this S = 0.034 mol L^-1 
MW of pbBr₂ = 367 
$\therefore$Solubility S = 0.034 mol L^-1 =0.034 x 367 gL^-1 = 12.48 
12.48 = P + 3.48 
$\therefore$P =12.48 – 3.48 = 9