Note: The 10 Sunsworth figure was wrong for this particular pulsar but is realistic in general. It is not the luminosity of the pulsar. it is the "thermal luminosity" of the pulsar's orbiting star.
The luminosity for pulsars in general ranges from the order of $10^{29}$ ergs/second to the Crab pulsar which is $10^{38}$ ergs/second. The Sun is about $10^{33}$ ergs/second. So we see pulsars range from $0.0001$ to $100000$ Sunsworth. 10 Sunsworth falls in that range. But the range is so big you can really make the pulsar as powerful as you like.
The figures on the table are not the exact measured luminosity. They are the so-called "spin-down luminosity". This is a theoretical upper bound based on the period and mass of the pulsar, and how quickly it is losing kinetic energy.
520 Sunsworth
If you stand on the planet then 520 times as much energy hits you from the pulsar jet, compared to how much would hit you from the Sun on Earth.
HOWEVER: The beam will hit the planet for a very brief amount of time before it rotates out of focus again. So you will not have time to be cooked alive. It is an exercise for the reader to compute if you will be cooked alive over several rotations.
The pulsar PSR J0952–0607 weighs about 3 Suns. The luminosity is about 10 Suns' worth. The following method works the same if you substitute figures from your favorite pulsar.
Luminosity is the amount of energy released by the star. For a normal star the energy is radiated symmetrically. For a pulsar the energy goes out the two jets. So each jet shoots out 5 Suns worth of energy.
For simplicity let's say the emission shoots out in a cone of 10 degrees. That's consistent with this answer on Physics Exchange
For simplicity we'll suppose your planet is exactly 1 AU from the pulsar. Consider the sphere of all points 1 AU from the pulsar. The jet sprays into "some spherical cap" (the blue part of the sphere surface)

If that cap is big then the energy is more spread out and less hits the planet. If the cap is small then the beam is focused and the planet gets more energy.
Fortunately the area is easy to compute. The formula is $A = S (1-\cos \theta)$ for $S$ the area of the whole sphere. For $\theta = 10^\circ$ we have $\cos \theta =0.985$ and $A = 0.015 S$.
That means we have 5 Sunsworth of energy compressed into an area $0.015$ the size of the sphere. So the density of the energy passing through the cap is $5$ times $1/0.015 \simeq 66.666\ldots $ times the energy density of the Sun on the Earth. The total is $5/0.015 = 333.333$ Sunsworth.
Move the planet to 0.8 AU and the area of the cap shrinks by $0.8^2 = 0.64 $. So multiply the density by $1/0.64 = 1.5625$ to get about $520$ Sunsworth.
If you shrink the angle to 1 degree as mentioned in the same Physics answer, then we have $\cos \theta = 0.9998$ and the final answer goes up by a factor of 100. So it would be
52,000 Sunsworth.