Boolean expression to NOR-gates
So Complement Law says \$\overline{\overline{X}} = X\$ 

We start with AND - OR.
$$BD + \overline{A}B\overline{C} + \overline{A}CD$$
Double Complement.
$$\overline{\overline{BD + \overline{A}B\overline{C} + \overline{A}CD}}$$
Use DeMorgan's Theorem to remove lower complement.
$$\overline{\overline{BD} • \overline{\overline{A}B\overline{C}} • \overline{\overline{A}CD}}$$
AND - OR has become NAND - NAND.  Use DeMorgan's on terms.
$$\overline{(\overline{B} + \overline{D}) • (A + \overline {B} + C) • (A + \overline{C} + \overline{D})}$$
NAND - NAND has become OR - NAND.  Use DeMorgan's Theorem to remove complement.
$$\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}$$
OR - NAND has become NOR - OR.  Double Complement again.
$$\overline{\overline{\overline{(\overline{B} + \overline{D})} + \overline{(A + \overline {B} + C)} + \overline{(A + \overline{C} + \overline{D})}}}$$
NOR - OR become NOR - NOR.  With an extra NOR connected as a NOT gate.