Given,

Two dices are thrown so total sample space will be $36$,

Now let the sum of numbers appearing on dice be a multiple of $4$ be event $A$,

So outcomes of event $A$ will be $\left\{\left(2,2\right),\left(1,3\right),\left(3,1\right),\left(5,3\right),\left(3,5\right),\left(6,2\right),\left(2,6\right),\left(4,4\right),\left(6,6\right)\right\}\to 9$

So probability will be $P\left(A\right)=\frac{9}{36}=\frac{1}{4}$

Now let the event $4$ has already occured be $B$,

So outcomes will be $\left\{\left(1,4\right),\left(4,1\right),\left(2,4\right),\left(4,2\right),\left(3,4\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(5,4\right),\left(4,6\right),\left(6,4\right)\right\}\to 11$

So probability will be $P\left(B\right)=\frac{11}{36}$

So $P\left(A\cap B\right)=\frac{1}{36}$

Now $P\left(B|A\right)=\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}}{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}}=\frac{1}{9}$

So, $3p+2=3\times \frac{1}{9}+2=\frac{7}{3}$