Getting Equivalent Resistance WIth delta and star transforms
You don't need \$\Delta-Y\$ transformations to solve that circuit. Consider the topmost triangle, and divide its base in a series of two \$4.5\,\mathrm{k\Omega}\$ resistors. You obtain the following circuit (I rearranged the elements with respect to the original circuit): 







simulate this circuit &ndash; Schematic created using CircuitLab

The above circuit is symmetric with respect to the axis crossing the nodes c,d and e: all the resistances on the left are equal to those on the right. This implies that nodes c,d and e are at the same potential: you can short them together without altering the currents flowing in the circuit branches. The simplified circuit is the following:







simulate this circuit

You can see that now the network is composed only of series and parallel connections, and since it's symmetric, you can calculate the resistance of just one side and multiply it by two.

You can see that \$R_1\$, \$R_3\$ and \$R_7\$ are in parallel, for an equivalent resistance of \$2.25\,\mathrm{k\Omega}\$. The total resistance is thus

$$R_\mathrm{ab} = 2\times\frac{(2.25\,\mathrm{k\Omega}+9\,\mathrm{k\Omega})\times 9\,\mathrm{k\Omega}}{2.25\,\mathrm{k\Omega}+9\,\mathrm{k\Omega}+ 9\,\mathrm{k\Omega}} = 10\,\mathrm{k\Omega}$$.