Assume that the link AB makes an angle $a[t]$ with the horizontal and the link CB makes an angle $c[t]$ with the horizontal.
Equating the position of B (x and y coordinates) computed w.r.t both A and C, we get two equations.
$$ 250 \cos (a(t))+200=250 \cos (c(t)) $$
$$ 250 \sin (c(t))+150=250 \sin (a(t)) $$
These equations can be solved for $a[t]$ and $c[t]$.
$$ a(t)=\tan ^{-1}\left(\frac{3+4 \sqrt{3}}{3 \sqrt{3}-4}\right),\ c(t)=\tan
^{-1}\left(\frac{4 \sqrt{3}-3}{4+3 \sqrt{3}}\right)$$
The velocity of B (x and y coordinates) can also be computed in two ways. So we get two more equations.
$$ -250 c'(t) \sin (c(t))=-250 a'(t) \sin (a(t))$$
$$ 250 c'(t) \cos (c(t))+\text{vc}=250 a'(t) \cos (a(t))+\text{va}$$
These equations can be solved for $a'[t]$ and $c'[t]$.
$$ a'(t)=-\frac{\left(\sqrt{3}-4\right)
(\text{va}-\text{vc})}{1250},\ c'(t)=\frac{\left(4+\sqrt{3}\right)
(\text{va}-\text{vc})}{1250}$$
The velocity of B can now be computed as
$$ \left\{-250 c'(t) \sin (c(t)),250 c'(t) \cos (c(t))+\text{vc}\right\}$$
or
$$ \left\{-250 a'(t) \sin (a(t)),250 a'(t) \cos (a(t))+\text{va}\right\} $$
Substituting the values of $a[t]$, $c[t]$, $a'[t]$, $c'[t]$ and $va=5$, $vc=-3$ gives the same result in both cases.
$$ \left\{-3.60, 5.43\right\} $$
The point B is moving upwards and to the left.