Determine reactions at supports:

$(2$kN

Step $1$: Draw FBD at pulley and solve for forces in each cable

Step $2$: Draw FBD of member ABC and solve for reactions at A and $B.$

Pin at A resists two components: FAx and FAy

Since member BD is a two$-$force member, pin at B can be replaced with FBD along member BD

Left cable of the pulley at $C$ is replaced with $6$ kN force facing down

Right cable of the pulley at $C$ is replaced with $6$ kN force facing down AND $6$ kN force facing left. Note that since radius of the pulley is $0\mathrm{.}3$ m $,$ force facing down is $0\mathrm{.}3$ m to the right of point $C$