What is the electric potential at point P as shown in the figure? Assume that rod has uniform charge `lambda` per unit length.

Since the nothing on the attached image is given about the location of point P, I will assume the general case that point P is anywhere between the charges rods and label its coordinates as  `(x_P, y_P)` . (Let's assume the coordinate system has origin at the point where the rods meet and the axis coincide with the rods.)

Since electric potential is scalar, the potential at point P will be the scalar sum of the potential due to the field of each rod. Let's find the potential due to the horizontal rod first.

The electric potential due to the point charge can be found as `V = (kq)/r`  , where k is a constant, q is the value of point charge, and r is the distance to the point charge.

The charged rod can be thought of the collection of the point, or differential charges dq such that `dq = lambda*dx`  . The location of each differential charge on the horizontal rod is (x, 0). According to the distance formula, the distance between point P and each point charge is `sqrt((x-x_P)^2 + y_P^2)`.

Then the differential potential due to each differential charge is

`dV = (klambdadx)/sqrt((x-x_P)^2 + y_P ^2)`

and the potential due to the semi-infinite horizontal rod is

`V = int dV = int_0 ^oo (klambdadx)/sqrt((x-x_P)^2 + y_P^2)`

To take this integral, I will first use substitution `z = x-x_P`  . This will result in the change of the lower limit from 0 to `-x_P`  . The upper limits remains infinity and dz = dx.

The resultant integral can be looked up in the table or taken using trigonometric substitution. The result will be

`V = klambda arctan(z/y_P) |_(-x_P) ^oo`

At infinity, arctangent approaches `pi/2` , so

`V = klambda(pi/2 +arctan(x_P/y_P))`

The procedure for finding the potential due to the vertical rod is identical, except that x and y are reversed. The potential due to the vertical rod is

`V = klambda(pi/2 + arctan(y_P/x_P))`

When adding the potentials, notice that that the sum involves arctangent of a number plus arctangent of the reciprocal of that number. The sum of these two arctangents is `pi/2` : `arctan(t) + arctan(1/t) = pi/2` 

So the potential at point P due to both rods is

`klambda(pi/2 + pi/2 + pi/2) = (3pi)/2 klambda` .