Why is a ‘Voltage Drop’ considered when a current is passed through a resistance? A resistance, by definition offers restriction to the flow of current. Thus when propelling a charge through a resistance, we have to do more work. Therefore shouldn’t the work done per unit charge, i.e, voltage, increase?
 A: No, the voltage drop is a term for the energy spent in pushing the charge through.
If you do more work, then you push that charge through faster. Energy is still lost to heat etc. as it passes through the resistor. That is what we call the voltage drop
(Remember that voltage is a word for difference in electric potential energy per charge, so we are just talking about a certain energy amount (per charge).)
Think of voltage as the pressure difference in a water hose. If you squeeze it on the middle, then you must open up the faucet a little more to increase the pressure if you still want equally much water to come out. That is the extra work you do. The drop in the pressure as this water passes the constriction represents the voltage drop (on the other side there is less pressure, corresponding to less energy associated with the charge).
A: 
shouldn’t the work done per unit charge, i.e, voltage, increase?

The work done by the E field is $W=\int \vec F \cdot d \vec s$. And since $\vec E=\vec F/q$ we have $W/q=\int \vec E \cdot d\vec s$ but $\Delta V = -\int \vec E \cdot d\vec s$ so $\Delta V = -W/q$
This is sometimes described as the work done against the field rather than the work done by the field. Work done per unit charge is unfortunately too vague to determine the sign of the work. It does not describe whether it is seen from the perspective of the circuit or the perspective of the environment.
Unfortunately, this is a common confusion. I think it is therefore best to simply remember the completely unambiguous $$\Delta V =-\int \vec E \cdot d \vec s$$
A: A higher resistance requires more energy, at constant current, but this energy is converted into heat.
