Answer
W is not a vector space.
Work Step by Step
$W\subset \mathbb{R}^{4}$.
In order to be a subspace of $\mathbb{R}^{3}$ (definition, p.195) , W must:
(a) contain zero,
(b) be closed over addition
(c) be closed over multiplication by scalars.
Checking,
(a)
If $\left[\begin{array}{l}
0\\
0\\
0\\
0
\end{array}\right]$ = $\left[\begin{array}{l}
b-5d\\
2b\\
2d+1\\
d
\end{array}\right]$ , then
$2d+1=0$ and $d=0 \Rightarrow 1=0$,
which can not be,
so $\left[\begin{array}{l}
0\\
0\\
0\\
0
\end{array}\right]\not\in W$
Therefore, W is not a subspace of $\mathbb{R}^{4}$,
W is not a vector space.