Simple MOS current mirror bias circuit
You have a misconception at this point. That the transistor is diode connected only means that \$V_{ds} = V_{gs} = V_{bias}\$, but that's all! This by its own doesn't mean that \$V_{gs}\$ will be equal to \$V_{t}\$.

\$V_{t}\$ is just a threshold that should be at least achieved for the transistor to conduct. You shouldn't mix this with the case of a normal diode. For the transistor, we just say that in this configuration it's I-V characteristic "looks like" that of a diode, but it doesn't mean that \$V_{gs}\$ will remain equal to \$V_{t}\$ no matter how much current it flows. The "diode configuration" just makes sure that the MOSFET will be in saturation at all case, once it starts conducting since \$V_{ds}\$ aka \$V_{gs}\$ will always be higher than \$V_{gs}-V_t\$

What you actually achieve with the current mirror, is that by forcing the current through the transistor to be \$I_{bias}\$, you basically define how much \$V_{gs} = V_{bias}\$ will be. Take a look at the typical I-V characteristics of a MOSFET. You can see it like this: You define how much \$I_d\$ will be with your bias circuit above the MOSFET. Then draw a horizontal line at this value of \$I_d\$ and you find how much \$V_{gs}\$ will be forced to be.

So, all in all, with \$I_{bias}\$ having a specific and constant value set by your bias circuit, \$V_{bias}\$ will also have a specific and constant value that will be, normally, higher than \$V_{t}\$ and that depends on how much \$I_{bias}\$ is. In this way, your only free parameter to play with is indeed the Width of the transistors.