Start by drawing neat, correct, labeled free body diagrams for the lower handle and the lower jaw.  Recognize that piece $CE$ is a two force member in compression so the force on pin $E$ acts down and left along a line passing through $C$ and $E$.

Geometry

Determine angle $\theta$ from the geometry of the pliers.  

$$\theta = \tan^{-1}\left( \dfrac{\var{q(x3)}}{\var{q(y1)}}\right) = \var{d(theta)}°$$

FBD II

$$\begin{align}\text{II: } \Sigma M_B &= 0\\ CE_x \,(\var{q(y2)}) -CE_y\,(\var{q(x2)}) + P \,( \var{q(x2+x3+x4)})& = 0 \\ CE\left( \var{y2} \sin \theta - \var{x2} \cos \theta \right) &=  -P \,( \var{q(x2)}+\var{q(x3)}+\var{q(x4)})\\CE &=-P \,\left( \dfrac{\var{q(x2+x3+x4)}}{\var{-q(dperp)}} \right)\\ &= \var{d(ce)} P \\ \\ \text{II: }\Sigma F_x &= 0\\ B_x &= CE_x\\ &= CE \sin{\theta}\\ &= ( \var{d(ce)} P )\,(\sin \var{d(theta)}°)\\&= \var{d(bx)} P\end{align}$$

FBD I

$$\begin{align} \text{I: }\Sigma M_D &=0\\ A\,(\var{q(x1)}) &= B_x\,(\var{q(y1)} + \var{q(y2)})\\ A &= (\var{d(bx)} P) \, \left(\dfrac{\var{q(y1+y2)}}{\var{q(x1)}}\right)\\  &= \var{d(A)} P\end{align}$$